COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS

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1 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 8. The Multiplication Principle; Permutations Your Turn Each of the four digits can be one of the ten digits 0,,,, 0, so there are or 0,000 possible sequences. If no digit is repeated, there are 0 choices for the first place, 9 for the second, 8 for the third, and 7 for the fourth, so there are possible sequences. Your Turn Any of the 8 students could be first in line. Then there are 7 choices for the second spot, for the third, and so on, with only one student remaining for the last spot in line. There are ,0 different possible lineups. Your Turn The teacher has 8 ways to fill the first space (say the one on the left), 7 choices for the next book, and so on, leaving 4 choices for the last book on the right. So the number of possible arrangements is Your Turn 4 There are letters. If we use of the, the number of permutations is! 4 P(, ) ( - )! 4 0. Your Turn The word Tennessee contains 9 letters, consisting of t, 4 e s, n s and s s. Thus the number of possible arrangements is 9! 780.!4!!! Your Turn 7 The student has pairs of socks, so if the pairs were distinguishable there would be 4! possible selections for the next two weeks. But since the pairs of each color are identical, the number of distinguishable selections is 8. Exercises 4!,,0. 4!!!!.! ! ! ».08 0! ».09 0 Your Turn If the panel sits in a row, there are 4! or 4 ways of arranging the four class groups. Within the groups, there are! ways of arranging the freshmen,! ways of arranging the sophomores,! ways of arranging the juniors, and! ways of arranging the seniors. Using the multiplication principle, the number of ways of seating the panel with the classes together is 4!!!! 4...!! P (, ) ( - )!!!!!! P (, ) ( - )! 9! 0 9! 9! 0 Copyright 0 Pearson Education, Inc. 487

2 488 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS ! 8! P (8,7) (8-7)!!».04 0!! P (,9) ( - 9)! 4!» n! n! Pn (,0) ( n - 0)! n! n! n! n! Pnn (, ) n! ( n - n)! 0! n! n( n - )! Pn (,) n ( n -)! ( n -)! n! n! Pnn (, - ) [ n - ( n -)]! ( n - n + )! n! n!!. By the multiplication principle, there will be different home types available. 4. By the multiplication principle, there will be different meals possible.. There are 4 choices for the first name and choices for the middle name, so, by the multiplication principle, there are 4 0 possible arrangements.. The number of ways to choose a slate of officers is!! P (, ) ( - )!! 4!! There is exactly one -letter subset of the letters A, B, and C, namely A, B, and C. 0. In Example, there are only unordered -letter subsets of letters A, B, and C. They are AB, AC, and BC.. (a) initial This word contains i s, n, t, a, and. Use the formula for distinguishable permutations with n 7, n, n, n, n, and n. 4! 7!!!!!!!!!!! 7 4!! 840 n n n n n4 n There are 840 distinguishable permutations of the letters. (b) little Use the formula for distinguishable permutetions with n, n, n, n, and n 4.!!!!!!!! 4 80 There are 80 distinguishable permutations. (c) decreed Use the formula for distinguishable permutations with n 7, n, n, n, and n 4. 7! 7!!!!!!! 7 4!! 40 There are 40 distinguishable permutations.. Use the formula for distinguishable permutations. The number of different words is n!! 40,40. n! n! n! n4!!4!!!. (a) The 9 books can be arranged in (9, 9) 9!,880 ways. P (b) The blue books can be arranged in 4! ways, the green books can be arranged in! ways, and the red books can be arranged in! ways. There are! ways to choose the order of the groups of books. Therefore, using the multiplication principle, the number of possible arrangements is 4!!!! Copyright 0 Pearson Education, Inc.

3 Section (c) Use the formula for distinguishable permutations with n 9, n 4, n, and n. The number of distinguishable arrangements is 9! ! 4!!! 4! 0. (d) There are 4 choices for the blue book, for the green book, and for the red book. The total number of arrangements is 4 4. (e) From part (d) there are 4 ways to select a blue, red, and green book if the order does not matter. There! ways to choose the order. Using the multiplication principle, the number of possible ways is 4! (a) Since there are 4 distinguishable objects to be arranged, use permutations. The number of arrangements is P (4,4) 4! 87,78,9,000 0 or (b) There are! ways to arrange the pyramids among themselves, 4! ways to arrange the cubes, and 7! ways to arrange the spheres. We must also consider the number of ways to arrange the order of the three groups of shapes. This can be done in! ways. Using the multiplication principle, the number of arrangements is!4!7!! ,4,0. (c) In this case, all of the objects that are the same shape are indistinguishable. Use the formula for distinguishable permutations. The number of distinguishable arrangements is n! 4! 0,0. n! n! n!!4!7! (d) There are choices for the pyramid, 4 for the cube, and 7 for the sphere. The total number of ways is (e) From part (d) there are 84 ways if the order does not matter. There are! ways to choose the order. Using the multiplication principle, the number of possible ways is 84! ! 0 9! To find the value of 0!, multiply the value of 9! by 0.. 4! 4 40! 000» » (a) The number! has factors of five so there must be ending zeros in the answer. (b) The number 7! has factors of five (one each in, 0,, and 0 and two factors in ), so there must be ending zeros in the answer. (c) The number 7! has + 8 factors of five (one each in,0,,7 and two factors each in, 0, and 7), so there must be 8 ending zeros in the answer. 8. (a) Since! has two s, there are two ending zeros in the answer to!. Thus,! ¹ 479,00,0. (b) Since! Has four s, there are four ending zeros in the answer to!. Thus,! ¹,8,0,740,000,000,000,000. (c) Since! has three s, there are three ending zeros in the answer to!. Thus,! ¹,07,4,80,000. (d) Since 4! has two s, there are two ending zeros in the answer to 4!. Using a calculator, 4! is approximated at E0. Since the last two digits are zero, 4! 87,78, 9, ! 4! P (4, 4) (4-4)! 0! If0! 0, then P (4,4) would be undefined. 0. Use the multiplication principle. There are 4 (4)() (8)(7)()(48)()()().00 0 different possible bags.. (a) By the multiplication principle, since there are 7 pastas and sauces, the number of different bowls is 7 4. (b) If we exclude the two meat sauces there are 4 sauces left and the number of bowls is now Copyright 0 Pearson Education, Inc.

4 490 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS. Ranking investments out of 9 amounts to finding the number of permutations of 9 elements taken at a time, which is 9! 9! P (9, ),0. (9 - )! 4!. (a) Since there are slots, the commercials can be arranged in! 9,9,800 ways. (b) Use the multiplication principle. We can put either stores or restaurants first ( choices). Then there are! orders for the stores and! orders for the restaurants, so the number of groupings is!! 7,800. (c) Since the number of restaurants is one more than the number of stores, a restaurant must come first. This eliminates the first choice in part (b), but we still can order the restaurants and the stores freely within each category, so the answer is!! 8, of the drugs can be administered in!! P (, ) ( - )! 7! ! 7! 9, 040 different sequences.. If each species were to be assigned initials, since there are different letters in the alphabet, there could be 7,7 different -letter designations. This would not be enough. If 4 initials were used, the biologist could represent 4 4,97 different species, which is more than enough. Therefore, the biologist should use at least 4 initials.. (a) The total number of presentations is + +, so the presentations can be scheduled in! 479, 00, 00 orders. (b) If we group the presentations for a given subject together, these three groups can present in! orders, and within each group there are!,! and! possible orders. So the number of possible orders is!!!! 7,800. (c) There are only physics presentations, so there are just ways of selecting the first and last presentation. The remaining 0 presentations can come in any order, so the number of possible orders is 0! 7, 7, The number of ways to seat the people is!! P (, ) 0! A ballot would consist of a list of the candidates for office and a list of the candidates for office. The number of ways to list candidates for office is P (,)!. The number of ways to list candidates for office is P (,)! 70. There are two ways to choose which office goes first. By the multiplication principle, the number of different ballots is The number of possible batting orders is 9! 9! P (9, 9) (9-9)! 0!,,8,40 0» The number of ways to select the 4 officers is!! P (, 4) ( - 4)!! 4!!,, (a) The number of ways works can be arranged is P (, )! 0. (b) If one of the overtures must be chosen first, followed by arrangements of the 4 remaining pieces, then P(,) P(4,4) 4 48 is the number of ways the program can be arranged. 4. (a) Pick one of the traditional numbers followed by an arrangement of the remaining total of 7. The program can be arranged in P(, ) P(7, 7) 7!, 00 different ways. (b) Pick one of the original Cajun compositions to play last, preceded by an arrangement of the remaining total of 7. This program can be arranged in P(7, 7) P(,) 7!,0 different ways. Copyright 0 Pearson Education, Inc.

5 Section (a) There are 4 tasks to be performed in selecting 4 letters for the call letters. The first task may be done in ways, the second in, the third in 4, and the fourth in. By the multiplication principle, there will be 4 7,00 different call letter names possible. (b) With repeats possible, there will be or, call letter names possible. (c) To start with W or K, make no repeats, and end in R, there will be 4 04 possible call letter names. 44. (a) There are odd digits:,,, 7, and 9. There are 7 decisions to be made, one for each digit; there are choices for each digit. 7 Thus, 78, phone numbers are possible. (b) The first digit has 9 possibilities, since 0 is not allowed; the middle digits each have 0 choices; the last digit must be 0. Thus, there are ,000 possible phone numbers. (c) Solve as in part (b), except that the last two digits must be 0; therefore there are ,000 possible phone numbers. (d) There are no choices for the first three digits; thus, 4 0 0,000 phone numbers are possible. (e) The first digit cannot be 0; in the absence of repetitions there are 9 choices for the second digit, and the choices decrease by one for each subsequent digit. The result is ,0 phone numbers. 4. (a) Our number system has ten digits, which are through 9 and 0. There are tasks to be performed in selecting digits for the area code. The first task may be done in 8 ways, the second in, and the third in 0. By the multiplication principle, there will be different area codes possible. Copyright 0 Pearson Education, Inc. There are 7 tasks to be performed in selecting 7 digits for the telephone number. The first task may be done in 8 ways, and the other tasks may each be done in 0 ways. By the multiplication principle, there will be 8 0 8,000,000 different telephone numbers possible within each area code. (b) Some numbers, such as 9, 800, and 900, are reserved for special purposes and are therefore unavailable for use as area codes. (c) There are 8 choices for the first digit, since it cannot be 0 or. Since restrictions are eliminated for the second digit, there are 0 possibilities for each of the second and third digits. Thus, the total number of area codes would be Under the old system there were IP addresses available. Under the new system there are IP addresses available 47. (a) There were 0 7,7, 000 license plates possible that had letters followed by digits. (b) There were 0 7,7, 000 new license plates possible when plates were also issued having digits followed by letters. (c) The number of possible new license plates added by the format change in 980 was (0)(000)( ) 7,70, Since a social security number has 9 digits with no restrictions, there are 9 0,000,000,000 ( billion) different social security numbers. Yes, this is enough for every one of the 8 million people in the United States to have a social security number. 49. If there are no restrictions on the digits used, there would be 0 00, 000 different -digit zip codes possible. If the first digit is not allowed to be 0, there would be ,000 zip codes possible.

6 49 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 0. Since a zip code has nine digits with no restrictions, there are 9 0,000,000,000 different 9-digit zip codes.. There are possible identical shapes on each card. There are possible shapes for the identical shapes. There are possible colors. There are possible styles. Therefore, the total number of cards is 8.. Since a 0-sided die is rolled times, the number of possible games is 0 or games.. There are possible answers for the first question and possible answers for each of the 9 other questions. The number of possible objects is 9,7,84. 0 questions are not enough. 4. (a) The number of different circuits is P(9,9) since we do not count the city he is starting in. P (9, 9) 9!,880 is the number of different circuits. (b) He must check half of the circuits since, for each circuit, there is a corresponding one in the reverse order. Therefore, (,880) 8,440 circuits should be checked. (c) No, it would not be feasible.. (a) Since the starting seat is not counted, the number of arrangements is 7 P (9, 9) 9!».4 0. (b) Since the starting bead is not counted and the necklace can be flipped, the number of arrangements is P(4, 4) 4! 4,89,4, Combinations Your Turn Use the combination formula. Your Turn 0! C (0, 4) 0!4! Since the group of students contains either or 4 students out of, it can be selected in C(,) + C(, 4) ways.!! C(, ) + C(, 4) +!!!4! Your Turn (a) Use permutations. P (0, 4) (b) (c) (d) Use combinations.! C (, ) 4!! Use combinations. 8! C (8, ) 8!! Use combinations and permutations. First pick 4 rooms; this is an unordered selection:! C (, 4)!4! Now assign the patients to the rooms; this is an ordered selection: P (4, 4) 4! 4 The number of possible assignments is 4 0. Your Turn 4 The committee is an unordered selection. 0! C (0, ) 40 7!! If the selection includes assignment to one of the three offices we have an ordered selection. Your Turn P (0, ) There are C(4,) ways to select aces from the 4 aces in the deck and C(48,) ways to select the remaining cards from the 48 non-aces. Now use the multiplication principle. C(4, ) C(48, ) 7,9 0,77 Copyright 0 Pearson Education, Inc.

7 Section Warmup Exercises W. P (,) 4 0 W. There are a s and two n s, one b and one s, so the number of distinguishable permutations is 7! 40. (!)(!)(!)(!) n! Cn (,0) ( n - 0)!0! n! n! n! n! n! Cnn (, ) ( n - n)! n! 0! n! n! 8. Exercises. To evaluate C (8, ), use the formula Cnr (, ) n! ( n - r)! r! 9. Cn (,) n! ( n - )!! nn ( - )! ( n -)! n with n 8 and r. 8! C (8, ) (8 - )!! 8!!! 8 7!! 0. Cnn (, - ) n! [ n - ( n -)]!( n -)! n! ( n - n + )!( n -)! nn ( - )!!( n - )! n 4.!! C (, ) ( - )!! 7!! ! 7! There are clubs, from which are to be chosen. The number of ways in which a hand of clubs can be chosen is! C (, ) 7. 7!!. To evaluate C (44, 0), use the formula. Cnr (, ) n! ( n - r)! r! with n 44 and r 0. C (44, 0) 44! (44-0)!0! 44! 4!0! ! C (40,8) (40-8)!8! 40!!8!».4 0. There are red cards in a deck, so there are C(, ) ways to select a hand of red cards.. (a) There are! C (, ) 0, 0 0!!! 4! C(, ) 0!!! different -card combinations possible. (b) The 0 possible hands are {,},{,},{,4},{4,},{,}, {,4},{,},{,4},{,},{,}. Of these, 7 contain a card numbered less than. Copyright 0 Pearson Education, Inc.

8 494 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 4. (a) The number of ways to select a committee of 4 from a club with members is C (, 4), 4. (b) If the committee must have at least member and at most members, it must have,, or members. The number of committees is C(, ) + C(, ) + C(, ) Choose letters from {L, M, N}; order is important. (a) L M L M N L M N L N M N There are 9 ways to choose letters if repetition is allowed. (b) L M N M N L M There are ways to choose letters if no repeats are allowed. (c) The number of combinations of elements taken at a time is! C (, ).!! This answer differs from both parts (a) and (b).. (a) With repetition permitted, the tree diagram shows different pairs. N L L M N P L M N P L M N P L M N P L M N P (b) If repetition is not permitted, one branch is missing from each of the clusters of second branches, for a total of different pairs. L M N P M N P L N P L M P (c) Find the number of combinations of 4 elements taken at a time. C (4, ) No repetitions are allowed, so the answer cannot equal that for part (a). However, since order does not matter, our answer is only half of the answer for part (b). For example, LM and ML are distinct in (b) but not in (c). Thus, the answer differs from both (a) and (b). 7. Order does not matter in choosing members of a committee, so use combinations rather than permutations. (a) The number of committees whose members are all men is 9! C (9, ) 4!! 9 8 7! 4!. (b) The number of committees whose members are all women is L M N! C (, )!! !! 4 4. Copyright 0 Pearson Education, Inc.

9 Section (c) The men can be chosen in 9! 9 8 7! C(9, )!!! 84 ways. The women can be chosen in! 0 9! C(, ) 9!! 9! ways. Using the multiplication principle, a committee of men and women can be chosen in ways. 4! C (4, 9) 4 ways.!!. Since order does not matter, the answers are combinations.! 4! (a) C(, ) 0 4!! 4! 0 samples of marbles can be drawn. (b) C (, 4) samples of 4 marbles can be drawn. (c) Since there are 9 blue marbles in the bag, the number of samples containing blue marbles is C (9, ). 8. Since order is not important, the answers are combinations. (a) If there are at least 4 women, there will be either 4 women and man or women and no men. The number of such committees is C(, 4) C(9,) + C(, ) C(9, 0) (b) If there are no more than men, there will be either no men and women, man and 4 women, or men and women. The number of such committees is C(9, 0) C(,) + C(9, ) C(, 4) + C(9, ) C(, ) Order is important, so use permutations. The number of ways in which the children can find seats is!! P (, ) ( - )!!! 479,00, Order does not matter, so use combinations. (a) The students who will take part in the course can be chosen in 4! C (4, )!! 4!! 4 ways. (b) The 9 students who will not take part in the course can be chosen in Copyright 0 Pearson Education, Inc.. Since order does not matter, use combinations. (a) There are C (, ) 00 possible samples of apples. (b) There are C (7, ) possible samples of rotten apples. (c) There are C(7,) C (9, ) 97 possible samples with exactly rotten apple.. Since order does not matter, use combinations.! 4! (a) C(, ) 0!!! There are 0 possible samples with all black jelly beans. (b) There is only red jelly bean, so there are no samples in which all are red. (c) C (, ) There is sample with all yellow. (d) C(, ) C (,) 0 0 There are 0 samples with black and red. (e) C(, ) C (,) 0 0 There are 0 samples with black and yellow. (f) C(, ) C (,) There are samples with yellow and black. (g) There is only red jelly bean, so there are no samples containing red jelly beans. 4. Since order is important, use a permutation. The plants can be arranged in P (9,) 9 8 7,0 different ways.

10 49 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS. Show that Cnr (, ) Cnn (, - r). Work with each side of the equation separately. n! Cnr (, ) r!( n - r)! n! Cnn (, - r) ( n - r)![ n - ( n - r)]! n! ( n - r)! r! Since both results are the same, we have shown that Cnr (, ) Cnn (, - r).. (a) We will call the digit on the left the first digit. No fives: There are 8 choices for the first digit and 9 choices for the second and third digits, giving 8(9 ) possibilities. Two fives: If the non- digit goes in the first place, it can have one of eight values; if it goes in the second or third place it can have one of 9 values, so these are possibilites. Three fives: For fives there is only one possibility, namely the number. Thus out of the 900 numbers from 00 to 999, 900 8(9 ) ( ) contain exactly one. (b) Multiply the numbers of possibilities along each path in the tree diagram and add these products. ()(9)(9) + (8)()(9) + (8)(9)() 7. Use combinations since order does not matter. (a) First consider how many pairs of circles there are. This number is! C (, ).!4! Each pair intersects in two points. The total number of intersection points is 0. (b) The number of pairs of circles is n! Cn (,) ( n - )!! nn ( -)( n- )! ( n - )! nn ( - ). Each pair intersects in two points. The total number of points is nn ( ) nn ( ) There are 7 digits. The number of cases with the same number of dots on both sides is C (7,) 7. The number of cases with a different number of dots on each side is C (7, ) 7!. The!! total number of dominoes that can be formed is C(7,) + C(7, ) Since the assistants are assigned to different managers, this amounts to an ordered selection of from 8. P (8, ) Since order is not important, use combinations. 0! C (0, ),8,70 4!!. Order is important in arranging a schedule, so use permutations.! (a) P (, )! 70 0! She can arrange her schedule in 70 ways if she calls on all prospects.! (b) P (, 4) 0! She can arrange her schedule in 0 ways if she calls on only 4 of the prospects.. Since order is not important, use combinations. (a) Since workers are to be chosen from a group of 9, the number of possible delegations is C (9, ) 84. (b) Since a particular worker must be in the delegation, the first person can only be chosen in way. The two others must be Copyright 0 Pearson Education, Inc.

11 Section selected from the 8 workers who are not the worker who must be included. The number of different delegations is C(8, ) 8. (c) We must count those delegations with exactly woman ( woman and men), those with exactly women ( women and man), and those with women. The number of delegations including at least woman is C(4,) C(,) + C(4,) C(,) + C(4,) There are types of meat and types of extras. Order does not matter here, so use combinations. (a) There are C (,) ways to choose one type of meat and C (, ) ways to choose exactly three extras. By the multiplication principle, there are C(,) C (, ) 0 40 different ways to order a hamburger with exactly three extras. (b) There are C (, ) 0 different ways to choose exactly three extras. (c) At least five extras means extras or extras. There are C (,) different ways to choose exactly extras and C (,) ways to choose exactly extras, so there are C(, ) + C(, ) + 7 different ways to choose at least five extras. 4. There are no restrictions as to whether the scoops have to be different flavors. (a) The number of different double-scoop cones will be 9. (b) The number of different triple-scoop cones will be 9,79. (c) There are! C (, ) 4 9!! ways to make double-scoop cones with two different flavors. In addition, there would be ways to make double-scope cones with the same flavors. Therefore, double-scoop cones can be made if order doesn t matter. (d) There is one mono-flavor triple-scoop cone. For two-flavor cones, we choose one of flavors to be the single scoop, leaving 0 flavors for the non-unique scoops. Finally, there are C(,) ways of building a cone with three different flavors. Thus there are ()(0)(9) + ()(0) + 4! different triple-scoop cones if order doesn t matter.. Select 8 of the smokers and 8 of the nonsmokers; order does not matter in the group, so use combinations. There are C(, 8) C (, 8) 4,, 49,900 different ways to select the study group.. Since the plants are selected at random, that is, order does not matter, the answers are combinations. (a) She is selecting 4 plants out of plants. The number of ways in which this can be done is C (, 4) 0. (b) She is selecting of the wheat plants and of the other plants. The number of ways in which this can be done is C(, ) C (, ) Order does not matter in choosing a delegation, so use combinations. This committee has members. (a) There are 9! C (9, )!! 9 8 7!! 84 possible delegations. (b) To have all Democrats, the number of possible delegations is C (, ) 0. (c) To have Democrats and Republican, the number of possible delegations is C(, ) C (4,) Copyright 0 Pearson Education, Inc.

12 498 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS (d) We have previously calculated that there are 84 possible delegations, of which 0 consist of all Democrats. Those 0 delegations are the only ones with no Republicans, so the remaining delegations include at least one Republican. 8. Since order is important, use permutations. 0! 0! P (0, 4) 040 (0-4)!! different committees are possible. 9. Order does not matter in choosing the panel, so use combinations. 4! C (4, ) 4!! ! 4! 4,90 The publisher was wrong. There are 4,90 possible three judge panels. 40. Since order does not matter, use combinations.! C (, ) ( - )!!! 9!!,0,9,00 4. Since the cards are chosen at random, that is, order does not matter, the answers are combinations. (a) There are 4 queens and 48 cards that are not queens. The total number of hands is C(4, 4) C (48,) (b) Since there are face cards ( in each suit), there are 40 nonface cards. The number of ways to choose no face cards (all nonface cards) is 40! C (40, ) 8,008.!! (c) If there are exactly face cards, there will be nonface cards. The number of ways in which the face cards can be chosen is C (, ), while the number of ways in which the nonface cards can be chosen is C (40,). Using the multiplication principle, the number of ways to get this result is C(, ) C (40, ) 9880,080. (d) If there are at least face cards, there must be either face cards and nonface cards, face cards and nonface cards, 4 face cards and nonface card, or face cards. Use the multiplication principle as in part (c) to find the number of ways to obtain each of these possibilities. Then add these numbers. The total number of ways to get at least face cards is C(, ) C(40, ) + C(, ) C(40, ) + C(, 4) C(40, ) + C(, ) , ,00 + 9, , 7. (e) The number of ways to choose heart is C(, ), the number of ways to choose diamonds is C(, ), and the number of ways to choose clubs is C(, ). Using the multiplication principle, the number of ways to get this result is C(, ) C(, ) C (, ) , (a) List the possibilities for each suit.,, 7, 8, 9;,, 7, 8, 0;,, 7, 9, 0;,, 8, 9, 0;, 7, 8, 9, 0;, 7, 8, 9, 0 There are possibilities for each suit and there are 4 suits, so there are 4 4 possibilities. (b) There are cards of each suit from to 0. Select of the cards. There are 4 C(, ) 4 4 possibilities. 4. Since order does not matter, use combinations. good hitters: C(, ) C(4, ) good hitters: C(, ) C(4, 0) 0 0 The total number of ways is Since the hitters are being chosen at random, that is, order does not matter, the answers are combinations. (a) The coach will choose of the good hitters and of the 8 poor hitters. Using the multiplication principle, this can be done in C(, ) C (8, ) 0 ways. Copyright 0 Pearson Education, Inc.

13 Section (b) The coach will choose of the good hitters. This can be done in C (, ) 0 ways. (c) The coach must choose either good hitters and poor hitter or good hitters. Add the results from parts (a) and (b). This can be done in C(, ) C(8,) + C(,) 40 ways. 4. Since order does not matter, use combinations. (a) There are C (0, ),04 different ways to select of the orchids. (b) If special orchids must be included in the show, that leaves 8 orchids from which the other orchids for the show must be chosen. This can be done in C (8, ) 8 different ways. 4. In the lottery, different numbers are to be chosen from the 99 numbers. (a) There are 99! C (99, ) 9!!,0,9, different ways to choose numbers if order is not important. (b) There are 99! P (99, ) 9! 80, 78,04,0 different ways to choose numbers if order matters. 47. Since order is not important, use combinations. To pick of the winning numbers, we must also pick of the 9 losing numbers. Therefore, the number of ways to pick of the winning numbers is C(, ) C (9, ) (a) The number of ways to form the two committees assuming the nominating committee is formed first is C(9, 7) C (, ) 9,907, 9. (b) The number of ways to form the two committees assuming the public relations committee is formed first is C(9, ) C (4, 7) 9,907, 9. Copyright 0 Pearson Education, Inc. (c) The number of indistinguishable ways can be thought of as first choosing the 7 places occupied by the red shirts out of 9 possible places in the line up and then choosing the places occupied by yellow shirts out of the remaining places. The number of ways is C(9, 7) C (, ) 9,907, 9. This is the same calculation as in part (a). 49. (a) The number of different committees possible is C(, ) + C(, ) + C(, 4) + C(, ) (b) The total number of subsets is. The number of different committees possible is - C(,) - C(,0) Three distinct letters can be selected in C(, ) ways, and once they are selected there is only one way to arrange them in alphabetical order. There are ways to select the three digits. So the total number of plates is! 000, 00, 000.!!. (a) If the letters can be repeated, there are choices of letters, and there are 0 choices for the digit, giving 0, 089,7, 70 passwords. (b) For nonrepeating letters we have P(, ) 0 or,7,,000 passwords.. (a) There can be, 4,,,, or no toppings. The total number of possibilities for the first pizza is C(, ) + C(, 4) + C(, ) + C(, ) + C(, ) + C(, 0) The total number of possibilities for the toppings on two pizzas is 04 04, 048, 7. (b) In part (a), we found that if the order of the two pizzas matters, there are 04,048,7

14 00 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS possibilities. If we had a list of all of these possibilities and if the order of the pizzas doesn t matter, we must eliminate all of the possibilities that involve the same two pizzas. There are 04 such items on the list, one of each of the possibilities for one pizza. Therefore, the number of items on the list that have a duplicate is, 048, 7-04, 047,. To eliminate duplicates, we eliminate the second listing of each of these, that is,, 047,, 77. Subtracting this from the number of possibilities on the list, we see that if the order of the two pizzas doesn t matter, the number of possibilities is, 048, 7 -, 77 4,800.. (a) A pizza can have,,, or no toppings. The number of possibilities is C(7, ) + C(7, ) + C(7, ) + C(7, 0) There are also four speciality pizzas, so the number of different pizzas is (b) The number of 4forAll Pizza possibilities if all four pizzas are different is C (88, 4) 0, 400,978, 0. The number of 4forAll Pizza possibilities if there are three different pizzas ( pizzas are the same and the other are different) is 88 C(87, ) 88 49,8 9,87, 708. The number of 4forAll Pizza possibilities if there are two different pizzas ( pizzas are the same or pizzas and pizzas are the same) is C(88, ) 70, ,70,0,09. The number of 4forAll Pizza possibilities if all four are the same is 88. The total number of 4forAll Pizza possibilities is 0, 400, 978, 0 + 9,87, 708 +, 0, ,9, 8,70 (c) Using the described method, there would be 87 vertical lines and 4 X s or 84 objects, so the total number is C (84, 4) 0, 9, 8, (a) C(8,) + C(8, ) + C(8, ) + C(8, 4) + C(8, ) + C(8, ) + C(8, 7) + C(8, 8) There are breakfasts that can be made. An easier way to compute this number is to notice that we could assemble a breakfast by deciding, for each of the 8 items, whether to 8 include it. This gives or breakfasts. But one of these is the empty breakfast containing no items. Since no one wants to eat this breakfast, we discard it and are left with - breakfasts. (b) She has choices. For the first choice she has 4 items. For the second choice she has 4 items. 4 4 She can make breakfasts. (c) She has C (4, ) choices of cereal mix and C (4, ) choices of add-in mix. Her total number of choices is C(4, ) C (4, ) 4 4. (d) He has C(4,) + C(4,) + C(4,) + C(4,4) choices of cereal mix and C(4,) + C(4,) + C(4,) + C(4,4) choices of add-in mix. His total number of breakfasts is. (e) C(7, 0) + C(7,) + C(7, ) + C(7, ) + C(7, 4) + C(7, ) + C(7, ) + C(7, 7) She has 8 different cereals. Copyright 0 Pearson Education, Inc.

15 Section !. (a) C (9, ) 84!! (b) ! (d) C (7, ) 80 4!! (e) First pick the two boneless buffalo wing flavors; there are C (, ) 0 ways of doing this. Then we still have 7 non-wing options, plus the buffalo chicken wings available for the third item. So our total is Consider the first conference with five teams in each of the three divisions. The three winners from the three divisions can result in, or, ways. Then, of the remaining - teams, the three wild card teams can be chosen in 0 ways. And since the order of the teams is not relevant, there are 0 ways to choose the teams from the first conference. In the second conference, the situation is the same with the single exception being that one division has six, not five, teams. Therefore, there are ways to choose the three division winners and ways to choose the wild card teams. Therefore, there are. ways to choose the six teams from the second conference. Finally, the number of ways the six teams can be selected from the first conference and the six teams can be selected from the second conference is 0,79, 70, (a) The number of ways the names can be arranged is 8!» (b) 4 lines consist of a syllable name repeated, followed by a syllable name and then a 4 syllable name. Including order, the number of arrangements is ,90,040. lines consist of a syllable name repeated, followed by two more syllable names. Including order, the number of arrangements is The number of ways the similar 4 lines can be arranged among the total lines is C (, 4). The number of arrangements that fit the pattern is 0,90,040 70» (a) The number of ways the judges can be selected is C (, 9) 0. (b) The number of different sets of scores is C(, 9) C (, 9) 0 48, Probability Applications of Counting Principles Your Turn Using method, C(, ) C(, ) P( NY, Chicago). C(, ) Your Turn Since 8 of the nurses are men, of them are women. We choose 4 nurses, men and women. Your Turn P( men among 4 selected) C(8, ) C(4, ) 48» 0.48 C(, 4) 7 The probability that the container will be shipped is the probability of selecting working engines for testing when there are working engines in the container. This is C(8, ) 4 P(all work). C(, ) 0 The probability that at least one defective engine is in the batch is 4 4 P (at least one defective) -» Your Turn 4 C(4, ) C(4, ) C(44, ) P( aces, kings, other) C(, ) 44,98,90» Copyright 0 Pearson Education, Inc.

16 0 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS Your Turn If the slips are chosen without replacement, there are P (7, ) 7 0 ordered selections. Only one of these spells now so P (now). If the slips are 0 chosen with replacement there are 7 choices for each and thus selections. Again, only one of these spells now so in this case P (now). 4 Your Turn Using method we compute the number of ways to arrange 4 pieces of fruit which come in 4 kinds: kiwis, apricots, 4 pineapples and coconuts. Assuming the pieces of each kind of fruit are indistinguishable (all kiwis look alike, and so on), the number of arrangements is 4!!!4!!,,0. If we keep the four kinds together (all kiwis next to each other, and so on) there are 4! 4 arrangements of the kinds that keep them together. So 4 P(all of same kind together),,0 -» Warmup Exercises W. There are 4 choices for the suit and then C(, ) ways of choosing from this suit, so there are (4) 84 4 ways of choosing the cards. W. C(9,) C(4,) 4,4, Exercises. There are C (, ) samples of apples. 0 9 C(, ) There are C (7, ) samples of red apples. 7 C(7, ) Thus, 7 P (all red apples).. There are C (, ) ways to select of the apples, while there are C (4,) 4 ways to select yellow ones. Hence, C(4, ) 4 P(yellow). C(, ). There are C (4,) samples of yellow apples. 4 C(4, ) There are C (7,) 7 samples of red apple. Thus, there are 7 4 samples of in which are yellow and red. Thus, 4 4 P ( yellow andred apple). 4. More red than yellow means or red. There are C (7, ) ways to choose red apples and C (4,) ways to pick a yellow; hence, there are C(7, ) C (4,) 84 ways to choose red. Since there are C (7, ) ways to pick red, we have ways to have more red than yellow. Therefore, 9 9 P(more red). C(, ) For Exercises through 0 the number of possible - member committees is C (0, ), C(9, ) P(all men) C(0, )» , 04 C(, ) P(all women) C(0, ) 4» , 04 C(9, ) C(, ) P( men, women) C(0, ) 40» 0.980, 04 C(9, ) C(, ) P( men, women) C(0, ) 940» 0.8, P(at least 4 women) P(4 women) + P( women) C(, 4) C(9, ) + C(, ) C(9, 0) C(0, )» 0.4, 04 Copyright 0 Pearson Education, Inc.

17 Section P(no more than men) P(no men) + P( man) + P( men) C(9, 0) C(, ) + C(9,) C(, 4) + C(9,) C(, ) C(0, ) 97» 0.04,04. The number of -card hands is C(, ).. There are C (4, ) ways to pick aces out of C (, ) ways to pick cards; hence, C(4,) P( aces) C(, )» There are C (, ) different -card hands. The number of -card hands with exactly one ace is C(4, ) C (48, ) The number of -card hands with two aces is C (4, ). Thus there are 98 hands with at least one ace. Therefore, P(the -card hand contains an ace) 98» There are C (, ) 78 ways to pick spades; hence, 78 P ( spades)» There are C(, ) different -card hands. There are C(, ) 78 ways to get a -card hand where both cards are of a single named suit, but there are 4 suits to choose from. Thus, P(two cards of same suit) 4 C(,) 4» 0.. C(, ) 7. There are C (, ) ways to pick face cards; hence, P ( face cards)» There are C(, ) different -card hands. There are face cards in a deck, so there are 40 cards that are not face cards. Thus, P(no face cards) C(40, ) 780 0» C(, ) 8. Ace,,, 4,,, 7, and 8 are the cards in each suit that are not higher than 8, for a total of, so P(no card higher than 8) C(, ) 49 48» C(, ) 9. There are choices for each slip pulled out, and there are slips pulled out, so there are,88,7 different words that can be formed from the letters. If the word must be chuck, there is only one choice for each of the letters (the first slip must contain a c, the second an h, and so on). Thus, P(word is chuck ) æ ö » è 0. Only the first letter is specified; the other 4 can be any letter. The probability of starting with the letter p is 0.08.». There are,88, 7 different words that can be formed. If the word is to have no repetition of letters, then there are choices for the first letter, but only choices for the second (since the letters must all be different), 4 choices for the third, and so on. Thus, P(all different letters) , 00 4, 97 8, 97» , Copyright 0 Pearson Education, Inc.

18 04 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS. There are possible -letter words, and words that do not contain x, y, or z. Hence, P(no x, y,or z),4,4» 0.47.,88, 7. P(at least presidents have the same birthday) - P(no presidents have the same birthday) The number of ways that 4 people can have 4 the same or different birthdays is (). The number of ways that 4 people can have all different birthdays is the number of permutations of things taken 4 at a time or P (, 4). Thus, P(at least presidents have the same birthday) P(, 4) -. 4 (Be careful to realize that the symbol P is sometimes used to indicate permutations and sometimes used to indicate probability; in this solution, the symbol is used both ways.). Using the result from Example, the probability that at least people in a group of n people have the same birthday is P(, n) -. () n Therefore, the probability that at least of the 00 U.S. Senators have the same birthday is P(, 00) () 7. Since there are 4 members of the House of Representatives, and there are only days in a year, it is a certain event that at least people will have the same birthday. Thus, P (at least members have the same birthday). 8. There are Cn (,) ways to pick which pair is to have the same birthday. One member of the pair has choices of a birthday, the other only. The other n - people have 4,,, etc., choices. Thus, the probability is P(, n - ) Cn (,). () n Copyright 0 Pearson Education, Inc. 9. P(matched pair) P( black or brown or blue) P( black) + P( brown) + ( blue) C(9,) C(,) C(,) + + C(7, ) C(7, ) C(7, ) The number of orders of the three types of birds is P(, ). The number of arrangements of the crows is P(, ), of the bluejays is P(4, 4), and of the starlings is P(, ). The total number of arrangements of all the birds is P(, ). P(all birds of same type are sitting together) P(, ) P(, ) P(4, 4) P(, ) P(, ) -4». 0. There are letters so the number of possible spellings (counting duplicates) is! 70. Since the letter is repeated times and the letter t is repeated times, the spelling little will occur!! 4 times. The probability that little will be spelled is There are letters so the number of possible spellings (counting duplicates) is! 9,9,800. Since the letter s is repeated 4 times, the letter s is repeated 4 times, and the letter p is repeated times, the spelling Mississippi will occur 4!4!! times. The probability that Mississippi will be spelled is ,9,800». Each of the 4 people can choose to get off at any 4 one of the 7 floors, so there are 7 ways the four people can leave the elevator. The number of ways the people can leave at different floors is the number of permutations of 7 things (floors) taken 4 at a time or P (7, 4) The probability that no passengers leave at the same floor is P(7, 4) 840» Thus, the probability that at least passengers leave at the same floor is (Note the similarity of this problem and the birthday problem. )

19 Section Let x the total number of balls. Since the probability of picking balls which all are blue is, we can see that x >. (If, x the probability would be.) Let s look at the number of blue balls needed. If there were blue balls, C (, ) and Cx (,), since the probability is. Since x must be larger than the number of blue balls, x ³ 7. But since C (7, ), Cx (,) ³ ¹. If there were 7 blue balls, C (7, ) and Cx (,) 4. Since x ³ 8, Cx (,) ³ ¹ 4. If there were 8 blue balls, C (8, ) and Cx (, ). Since x ³ 9, Cx (, ) ³ ¹. 8. There are C (, ) ways to choose typewriters.! 0 9 C(, )! 8! There are C (9, ) ways to choose nondefective typewriters. 9! C(9, ) 84!! Thus, P( drawn from the 9 are nondefective) There are C (9, 4) possible ways to choose 4 nondefective typewriters out of the C (, 4) possible ways of choosing any 4. Thus, C(9, 4) P(no defective). C(, 4) 0 If there were 9 blue balls, C (9, ) and Cx (, ). Since x ³ 0, Cx (,) ³, and x must be 0. Therefore, there were 0 balls, 9 of them blue, C(9, ) P(all blue). C(0, ). P(at least one $00-bill) P( $00-bill) + P( $00-bills) C(,) C(4,) C(, ) C(4, 0) + C(, ) C(, ) C(,0) C(4,) P(no $00-bill) C(, ) It is more likely to get at least one $00-bill.. There are ways to choose typewriter from the shipment of. Since of the are defective, there are 9 ways to choose nondefective typewriter. Thus, 9 P ( drawn from the is not defective). 7. There are C (9, ) possible ways to choose nondefective typewriters out of the C (, ) possible ways of choosing any. Thus, C(9, ) P(no defective). C(, ) Copyright 0 Pearson Education, Inc. 40. There are C(, 4) 49 different ways to choose 4 engines for testing from the crate of. A crate will not be shipped if any one of the 4 in the sample is defective. If there are defectives in the crate, then there are C(0, 4) 0 ways of choosing a sample with no defectives. Thus, P(shipping a crate with defectives) 0 4» There are C (, ) 79 ways to pick a sample of. It will be shipped if all are good. There are C (0, ) ways to pick good ones, so 7 P (all good)» If Melanie is first and Boyd last, the remaining 7 can be in any order. There are 9! possible orders, so 7! P (Melanie first, Boyd last). 9! 7 4. P(not Scottsdale customer) énumber of choices of 4 out of the ù non-scottsdale customers ê ú ë û number of choices of 4 out of the customers C(, 4) C(, 4) 44. There are P (, ) different orders of the names. Only one of these would be in alphabetical order. Therefore, P(, ) - are not in alphabetical order. Thus,

20 0 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS P(not in alphabetical order) P(, ) P(, ) There are 0 people in all, so the number of possible -person committees is C (0, ),04. Thus, in parts (a)-(g), ns ( ),04. (a) There are C (0, ) ways to choose the men and C (0, ) ways to choose the women. Thus, P( men and women) C(0, ) C(0, ) 0 4 C(0, ),04» (b) There are C (, ) ways to choose the Miwoks and C (9, ) ways to choose the Pomos. Thus, (c) P(exactly Miwoks and Pomos) C(, ) C(9, ) 0 C(0, ),04» Choose of the Miwoks, of the Hoopas, and of the 9 Pomos. Thus, P( Miwoks, Hoopas, and a Pomo) C(, ) C(, ) C(9,) 0 9 C(0, ),04» (d) There cannot be Miwoks, Hoopas, and Pomos, since only people are to be selected. Thus, ( Miwoks, Hoopas, and Pomos) 0. P (e) Since there are more women then men, there must be, 4, or women. P(more women than men) C(0, ) C(0, ) + C(0, 4) C(0, ) + C(0, ) C(0, 0) C(0, ) 77,04 (f) Choose of Hoopas and any of the non-hoopas. P(exactly Hoopas) C(, ) C(, ) C(0, ) 7» (g) There can be to Pomos, the rest chosen from the non-pomos. P(at least Pomos) C(9, ) C(, ) + C(9, ) C(, ) + C(9, 4) C(, ) + C(9, ) C(, 0) C(0, ) 0» (a) P (first person) 40 8 (b) (9!) P (last person) 40! (9!) 40(9!) 40 8 (c) No, everybody has the same chance. 47. There are books, so the number of selection of any books is C (, ) 4,4. (a) The probability that the selection consisted of Hughes and Morrison books is C(9, ) C(7, ) 8 C(, ) 4, 4 940» , 4 (b) A selection containing exactly 4 Baldwin books will contain of the books by the other authors, so the probability is (c) C(, 4) C(, ) 0 C(, ) 4, 4 00» , 4 The probability of a selection consisting of Hughes, Baldwin, and Morrison book is C(9, ) C(, ) C(7,) C(, ) 4, 4 0 4, 4» Copyright 0 Pearson Education, Inc.

21 Section (d) A selection consisting of at least 4 Hughes books may contain 4,, or Hughes books, with any remaining books by the other authors. Therefore, the probability is æc(9, 4) C(, ) C(9, ) C(, ) ö + C(9,) C(,0) è + C(, ) , , 4 99» , 4 (e) Since there are 9 Hughes books and Baldwin books, there are 4 books written by males. The probability of a selection with exactly 4 books written by males is C(4, ) C(7, ) 00 C(, ) 4, 4,0» , 4 (f) A selection with no more than books written by Baldwin may contain 0,, or books by Baldwin, with the remaining books by the other authors. Therefore, the probability is C(,0) C(,) + C(,) C(,) + C(,) C(,4) C(, ) , , , 00 4, 4 48,048» , There are C (,) different -card poker hands. There are 4 royal flushes, one for each suit. Thus, P(royal flush) 4 4 C(,),98,90 49,740 -» A straight flush could start with an ace,,, 4,, 7, 8, or 9. This gives 9 choices in each of 4 suits, so there are choices in all. Thus, P(straight flush) C(,),98,90» The four of a kind can be chosen in ways and then is matched with of the remaining 48 cards to make a -card hand containing four of a kind. Thus, there are 48 4 poker hands with four of a kind. It follows that P(four of a kind) 4 4 C(, ), 98, » A straight could start with an ace,,, 4,,, 7, 8, 9, or 0 as the low card, giving 40 choices. For each succeeding card, only the suit may be chosen. Thus, the number of straights is , 40. But this also counts the straight flushes, of which there are (see Exercise 49), and the 4 royal flushes. There are thus 0,00 straights that are not also flushes, so 0, 00 P (straight)» ,98,90. There are different values with 4 cards of each value. The total number of possible three of a kinds is then C(4,). The other cards must be chosen from the remaining 48 cards of different value. However, these cards must be different. Thus, for the last cards, there are 48 cards to choose from, but the cards must not have the same value. The number of possibilities for the last cards is C(48,) - C(4,), and P(three of a kind) C(4,) [ C(48,) - C(4,)] C(, )» There are different values of cards and 4 cards of each value. Choose values out of the for the values of the pairs. The number of ways to select the values is C(, ). The number of ways to select a pair for each value is C(4, ). There are cards that are neither of these values, so the number of ways to select the fifth card is C (44,). Thus, Copyright 0 Pearson Education, Inc.

22 08 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS C(, ) C(4, ) C(4, ) C(44,) P(two pairs) C(, ),» ,98,90 4. There are different values with 4 cards of each value. The total number of possible pairs is C(4, ). The remaining cards must be chosen from the 48 cards of different value. However, among these we cannot have of a kind nor can we have of a kind. P(one pair) C(4, )[ C(48, ) - C(4, ) - C(4, ) C(44,)] C(, )» 0.4. There are C (,) different -card bridge hands. Since there are only hearts, there is exactly one way to get a bridge hand containing only hearts. Thus, - P(only hearts)».7 0. C(,). The hand can have exactly aces or 4 aces. There are C (4,) 4 ways to pick exactly aces, and there are C (48,0) ways to pick the other 0 cards. Also, there is only C (4,4) way to pick 4 aces, and there are C (48,9) ways to pick the other 9 cards. Hence, C(4,) C(48,0) + C(4,4) C(48,9) P( aces) C(,)» There are C (4,) ways to obtain aces, C (4,) ways to obtain kings, and C (44,9) ways to obtain the remaining 9 cards. Thus, P(exactly aces and exactly kings) C(4, ) C(4, ) C(44, 9)» C(,) 8. The number of ways of choosing suits is P (4,). The number of ways of choosing of one suit is C (, ), 4 of another is C (, 4) and of another is C (, ). Thus, P( of one suit, 4 of another, and of another) P(4,) C(,) C(,4) C(,) C(,)» 0.0. Order is important in this problem because spades, 4 hearts, and clubs would be different than hearts, 4 clubs, and spades. For Exercises 9 through, use the fact that the number of 7-card selections is C (, 7). 9. Pick a kind for the pair: choices Choose suits out of the 4 suits for this kind: C (4, ) Then pick kinds out of the remaining: C (, ) For each of these kinds, pick one of the 4 suits: 4 The product of these factors gives the numerator and the denominator is C(, 7). C(4, ) C(, ) 4 C(, 7)» Pick the kinds for the pairs: C (, ) For each of these kinds, choose of the 4 suits: two factors of C (4, ) There are now cards remaining which must be of the kinds remaining: C (, ) For each of these kinds we pick one of the 4 suits: 4 The product of these factors gives the numerator and the denominator is C(, 7). C(, ) [ C(4, )] C(, ) 4 C(, 7)» 0.. Pick a kind for the three-of-a-kind: choices Choose suits out of the 4 suits for this kind: C (4, ) There are now 4 cards remaining which must be 4 of the kinds remaining: C (, 4) For each of these 4 kinds we pick one of the 4 4 suits: 4 The product of these factors gives the numerator and the denominator is C(, 7). 4 C(4, ) C(, 4) 4 C(, 7)» Pick a kind for the four-of-a-kind: choices Choose 4 suits out of the 4 suits for this kind: C (4, 4) (there s only one way of doing this!) There are now cards remaining which must be of the kinds remaining: C (, ) For each of these kinds we pick one of the 4 suits: 4 The product of these factors gives the numerator and the denominator is C(, 7). C(4, 4) C(, ) 4» C(, 7) Copyright 0 Pearson Education, Inc.

23 Section Pick a suit: 4 Now we either get exactly of this suit and of other suits: C(, ) C(9, ) or exactly of this suit and of another suit: C(, ) C(9,) or all 7 of our chosen suit: C (, 7). We now add these options over our usual denominator. 4 [ C(,) C(9, ) + C(, ) C(9,) + C(, 7)] C(, 7)» Pick kinds: C (, ) Pick either of the first kind and of the second and not of either of these kinds: C(4,) C(4,) C(44,) or pick of the first kind and of the second and not of either of these kinds: C(4,) C(4,) C(44,) or pick of each kind and of a different kind: C(4, ) C(4, ) C(44,) Noting that the first and second expressions above are the same, we can assemble the numerator over the usual denominator. C(,) [ C(4,) C(4,) C(44,) + C(4,) C(4,) C(44,)] C(,7)» We need at least hearts out of cards. This can happen in three ways: hearts, non-hearts: C(, ) C(9, ) 4 hearts, non-heart: C(, 4) C(9,) hearts: C (, ) Add these options over the usual denominator. C(,) C(9, ) + C(, 4) C(9,) + C(,) C(,7)» There are C (99, ),0,9, different ways to pick numbers from to 99, but there is only way to win; the numbers you pick must exactly match the winning numbers, without regard to order. Thus, P(win the big prize),0,9, -0» To find the probability of picking of the lottery numbers correctly, we must recall that the total number of ways to pick the lottery numbers is C(99, ),0,9,. To pick of the winning numbers, we must also pick of the 9 losing numbers. Therefore, the number of ways of picking of the winning numbers is C(, ) C (9,) 8. Thus, the probability of picking of the numbers correctly is C(,) C(9,) -7» C(99,) 8. Let A be the event of drawing four royal flushes in a row all in spades, and B be the event of meeting four strangers all with the same birthday. Then, 4 é ù PA ( ). ê C(, ) ú ë û For four people, the number of possible birthdays 4 is. Of these there are which are the same (that is, all birthdays January or January or January, etc.). PB ( ) 4 Therefore, 4 é ù PA ( Ç B) ê C(, ) ú ë û -4» No, this probability is much smaller than that of winning the lottery. 9. The probability of picking six numbers out of 49 is P( out of 49) C(49, ), 98,8 The probability of picking of picking five numbers out of is P( out of ) C(,),98,90 The probability of winning the lottery when picking five out of is higher. 70. (a) The number of ways to select numbers between and is C (,), 478, 7 and there are 4 ways to select the bonus number. P (winning jackpot), 478, 7 4 4,07, 9 (b) The number of selections you would make over 8 years is , 8, 480. The probability that none of the selections win is Copyright 0 Pearson Education, Inc.

24 0 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 7,48,480 æ4,07, 9 ö» è4,07, 9 Therefore, the probability of winning is This calculator answer is slightly inaccurate because of rounding in the calculator s arithmetic. The correct answer to four places is (a) The number of ways to select numbers between and 49 is C (49,),98,8. The number of ways to select of the numbers, while not selecting the bonus number is C(,) C (4,) 0,480 9, 00. The probability of winning fifth prize is 9, ,98,8» (b) The number of ways to select of the numbers plus the bonus number is C(, ) C(,) C (4, ), 480 7, 00. The probability of winning sixth prize is 7, ,98,8» 7. P(saying math class is tough ) C(,) C(9, )» C(70,4) No, it is not correct. The correct figure is.48%. 7. (a) There were 8 games played in the season, since the numbers in the Won column have a sum of 8 (and the numbers in the Lost column have a sum of 8). (b) Assuming no ties, each of the 8 games had possible outcomes; either Team A won and Team B lost, or else Team A lost and Team B won. By the multiplication principle, this means that there were (c) 8 8, 4, 4 different outcomes possible. Any one of the 8 teams could have been the one that won all of its games, any one of the remaining 7 teams could have been the one that won all but one of its games, and so on, until there is only one team left, and it is the one that lost all of its games. By the multiplication principle, this means that there were 8! ,0 different perfect progressions possible. (d) Thus, (e) 74. (a) (b) P( perfect progression in an 8-team league) 8! -4» If there are n teams in the league, then the Won column will begin with n -, followed by n -, then n -, and so on down to 0. It can be shown that the sum of nnso there ( ) these n numbers is, are ( )/ nn- different win/lose progressions possible. The n teams can be ordered in n! different ways, so there are n! different perfect progressions possible. Thus, P( perfect progression in an n-team league) n!. ( )/ nn - C(0, 9) C(, 9) C(48, 9) C(8, 9)» 9. 0» (a) There are only 4 ways to win in just 4 calls: the diagonals, the center column, and the center row. There are C (7,4) combinations of 4 numbers that can occur. The probability that a person will win bingo after just 4 numbers are called is 4 -».9 0. (b) C(7,4) There is only way to get an L. It can occur in as few as 9 calls. There are C (7,9) combinations of 9 numbers that can occur in 9 calls, so the probability of an L in 9 calls is -» C(7,9) (c) There is only way to get an X-out. It can ocur in as few as 8 calls. There are C (7,8) combinations of 8 numbers that can occur. The probability that an X-out occurs in 8 calls is -» C(7,8) Copyright 0 Pearson Education, Inc.

25 Section (a) (d) Four columns contain a permutation of numbers taken at a time. One column contains a permutation of numbers taken 4 at a time. The number of distinct cards is 4 P(, ) P(, 4)».4 0. æ ö P( blues drawn red removed) è and æ ö P( blues drawn blue removed) è. Since the box initially contained equal numbers of red and blue balls, P(red removed) P(blue removed). By Bayes Theorem, P(red removed blues drawn) ( blues drawn red removed) (red removed) ù ( blues drawn blue removed) (blue removed) úû [ P( blues drawn red removed) P(red removed) ] ép P êë + P P æö è ø æ ö æö + è ø è +» 0.99 (b) P(44 blues in 80 red removed) 44 æö æö C(80, 44) è è and P(44 blues in 80 blue removed) 44 æö æö C(80, 44) è è (See the next section for an explanation of this probability). By Bayes Theorem, P (red removed 44 blues in 80) [ P(44 blues in 80 red removed) P(red removed) ] ép(44 blues in 80 red removed) P(red removed) ù êë + P( 44 blues in 80 blue removed) P(blue removed) úû 44 æö æö C(80, 44) è è æ C(80, 44) ö æ ö + C(80, 44) æ ö æ ö è è è è» Binomial Probability Your Turn Your Turn 4 P(exactly of ) C(, )(0.9) (0.4) (0.48)(0.08)» 0.47 P(at least one incorrect charge) - P(no incorrect charges in 4) C(4, 0)(0.9) (0.7)» Your Turn P(at most incorrect charges in ) P(0) + P() + P() + P() 0 C(, 0)(0.9) (0.7) + C(,)(0.9) (0.7) + C(, )(0.9) (0.7) + C(, )(0.9) (0.7)» Warmup Exercises W. C (00,0)».7 0 W. C (, 8) 0,8, Exercises 4. This is a Bernoulli trial problem with P(success) P(girl). The probability of exactly x successes in n trials is Cnxp (, ) x (- p) n-x, where p is the probability of success in a single trial. We have n, x, and p Note that - p -. P(exactly girls and boys) æö æö C(, ) è è 0» 0. Copyright 0 Pearson Education, Inc.

26 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS. This is a Bernoulli trial problem with P(success) P(girl). The probability of exactly x successes in n trials is Cnxp (, ) x (- p) n-x, where p is the probability of success in a single trial. Here n, x, and p. P(exactly girls and boys) æö æö C(,) è è æ ö 0» 0. è. We have n, x 0, p, and - p. 0 æö æö P(no girls) C(,0) è è» We have n, x 0 is the number of boys, and P is the probability of having a boy. 0 æö æö P(no boys) C(,0)» 0.0 è èø. At least 4 girls means either 4 or girls. P(at least 4 girls) 4 0 æö æö æö æö C(, 4) + C(,) è èø èø èø +» We have, 4, or boys, so P(at least boys) æö æö æö æö C(,) + C(,4) è è è è æö æö + C(,) è è P (no more than boys) - P(at least 4 boys) - P(4boysorboys) -[ P(4 boys) + P( boys)] æ ö - + è - -» 0.8 P(no more than 4 girls) - P(girls) 0 æö æö - C(,) è è -» On one roll, P (). We have n, x, and p. Note that - p. Thus, 0 æö æö P(exactly ones) C(,) è è -0» We have n, x, and p, so. æö æö P(exactly ones ) C(,) è è ø» æö æö P(exactly one ) C(,) è è» 0.9. We have n, x, and p, so 0 æö æö P(exactly ones ) C(,) è è ø» Copyright 0 Pearson Education, Inc.

27 Section 8.4. No more than ones means 0,,, or ones. Thus, P(no more than ones) P(0 ones) + P( one) + P( ones) + P( ones) 0 æö æö æö æö C(, 0) + C(,) è è è è ø æö æö æö æö + C(, ) + C(, ) è è è è ø» P( incorrect charges) æ ö æ9ö C(,) è 0 è 0» 0.0 P(0 incorrect charges) 0 æ ö æ9ö C(, 0) è 0 è 0» No more than one means 0 one or one. Thus, P(no more than one) P(0 one) + P(one) æö æö æö æö C(,0) + C(,) è èø èø è» Cnr (,) + Cnr (, + ) n! n! + r!( n - r)! ( r + )![ n - ( r + ) ]! n!( r + ) r!( r + )( n - r)! n!( n - r) + ( r + )![ n - ( r + ) ]!( n - r) rn! + n! n( n! ) - rn! + ( r + )!( n - r)! ( r + )!( n - r)! rn! + n! + n( n! ) - rn! ( r + )!( n - r)! n! ( n + ) ( r + )!( n - r)! ( n + )! ( r + )![( n + ) - ( r + ) ]! Cn ( +, r+) 8. Since these two crib deaths cannot be assumed to be independent events, the use of binomial probabilities is not applicable and thus the probabilities that are computed are not correct. 9. Since the potential callers are not likely to have birthdates that are distributed evenly throughout the twentieth century, the use of binomial probabilities is not applicable and thus, the probabilities that are computed are not correct. For Exercises 0 through 4 we define a success to be the event that a customer is charged incorrectly. In this situation, n, p and - p P(at least incorrect charges) - P(no incorrect charges) (from Exercise )» P(at least incorrect charges) - P(0 or incorrect charges) 0 æ ö æ9ö - C(,0) è 0 è 0 ø» P(at most incorrect charges) 4 æ ö æ9ö - C(,) è 0 è 0 ø 0 4 æ ö æ 9ö æ ö æ9ö + C(, 0) C(,) è0 è 0 ø è0 ø è0 æ ö æ9ö + C(, ) è0 è 0» For Exercises through 8, we define a success to be the event that the family hardly ever pays off the balance. In this situation, n 0, p 0.4 and - p P(at least 4) 4 P() C(0, )(0.4) (0.74)» P(9) C(0, 9)(0.4) (0.74)» C(0, 0)(0.4) (0.74) C(0,)(0.4) (0.74)» C - C (0, )(0.4) (0.74) (0, )(0.4) (0.74) Copyright 0 Pearson Education, Inc.

28 4 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 8. P(at most ) 0 0 C(0, 0)(0.4) (0.74) 9 + C(0,)(0.4) (0.74) 8 + C(0, )(0.4) (0.74) 7 + C(0, )(0.4) (0.74) 4 + C(0, 4)(0.4) (0.74) + C(0, )(0.4) (0.74)» n 0, p 0.0, x P(0 defective transistors) C(0,0)(0.0) (0.9)» We have n 0, p 0.0, and - p 0.9. Thus, P(at most defective transistors) P(none defective) + P(one defective) + P(two defective) C(0,0)(0.0) (0.9) + C(0,)(0.0) (0.9) + C(0, )(0.0) (0.9)» Let success mean producing a defective item. Then we have n 7, p 0.0, and - p 0.9. (a) If there are exactly defective items, then. x Thus, P(exactly defective) C(7, )(0.0) (0.9)» (b) If there are no defective items, then x 0. Thus, (c) P(none defective) C(7, 0)(0.0) (0.9)» If there is at least defective item, then we are interested in x ³. We have. n 8, p 0.7 P(at least one defective) - Px ( 0)» (a) The probability that all 8 people like the product is (b) The probability that from 8 to 0 people (inclusive) like the product is P(exactly 8) + P(exactly 9) + P(exactly 0) C(8, 8)(0.7) (0.) + C(8, 9)(0.7) (0.) + C(8,0)(0.7) (0.)» (a) Since 80% of the good nuts are good, 0% of the good nuts are bad. Let s let success represent getting a bad nut. Then 0. is the probability of success in a single trial. The probability of 8 successes in 0 trials is C(0, 8)(0.) ( - 0.) C(0, 8)(0.) (0.8)» (b) Since 0% of the blowouts are good, 40% of the blowouts are bad. Let s let success represent getting a bad nut. Then 0.4 is the probability of success in a single trial. The probability of 8 successes in 0 trials is C(0,8)(0.4) ( - 0.4) C(0,8)(0.4) (0.)» (c) The probability that the nuts are blowouts is æprobability of Blowouts ö èhaving 8bad nuts out of 0ø æprobability of Good Nuts Blowouts ö or è having 8 bad nuts of éc(0, 8)(0.4) (0.) ù ê ú ë û éc(0, 8)(0.) (0.8) ù ê ú + 0. é 8 C(0, 8)(0.4) (0.) ù ë û êë úû» n 0, p 0.0 P(fewer than ) P(0) + P() + P() C(0,0)(0.0) (0.9) + C(0,)(0.0) (0.9) 8 + C(0, )(0.0) (0.9)» 0.94 The answer is e P(all 8) C(8, 8)(0.7) (0.)» Copyright 0 Pearson Education, Inc.

29 Section 8.4. n, p P(all ) C(,)(0.8) (0.)» n, p P(none) C(, 0)(0.8) (0.) -» n, p 0.8 P(not all) - P(all ) - C(,)(0.8) (0.)» n, p P(more than half ) P(8) + P(9) + P(0) + P() + P() + P() + P(4) + P() C(,8)(0.8) (0.) + C(,9)(0.8) (0.) C(,0)(0.8) (0.) + C(,)(0.8) (0.) + C(,)(0.8) (0.) + C(,)(0.8) (0.) C(,4)(0.8) (0.) + C(,)(0.8) (0.)» n 00, p 0.0, x P(exactly sets of twins) C(00, )(0.0) (0.988)» We have n 00, p 0.0, and - p Thus, P(more than half ) P( x 0) + P( x ) + P( x ) C(00,0)(0.0) (0.988) + C(00,)(0.0) (0.988) + C(00,)(0.0) (0.988) (0.0) (0.988) + 00(0.0) (0.0988) + 490(0.0) (0.988)» (a) (b) We distribute the 40 millon births that occur in 0 years evenly among the 700 hospitals, and then divide the births in any one hospital into strings of 9 consecutive births. Any one hospital has 40, 000, (700)(9) or about 9 such strings. Thus the probability of a boy-string in any one of the hospitals is 40, 000, ( ) (700)(9) (c) 4 or about 7 0. The probability of a boy-string in at least one hospital is minus the probability of no string in any hospital, which is ( ) 0.98 or about (d) Boys have a birth probability greater than (a) The probability of at least 4 male births out of 84 can be found using the binomial probability function on a calculator, and it is about The probability of at least 0 male births out of 4 is given by the calculator as (b) The male percentage is higher in the Clay County data, but the total number of births in Clay County is much smaller, so in fact the Macon County data is more surprising (less probable). (c) We are interested in the probability that the male number would be this high or higher, since any higher value also have been surprising. Any particular number of male births is very improbable, so the probability of exactly 4 male births is not a good measure of how surprising this event is. 4. n, p 0.04 (a) The probability that exactly men are colorblind is 48 P() C(, )(0.04) (0.98)» (b) The probability that no more than men are color-blind is Copyright 0 Pearson Education, Inc.

30 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS (c) P(no more than men are color-blind) 0 C(,0)(0.04) (0.98) + C(,)(0.04) (0.98) + C(, )(0.04) (0.98) 0 + C(,)(0.04) (0.98) C(, 4)(0.04) (0.98) 48 + C(,)(0.04) (0.98)» The probability that at least man is colorblind is - P(0 men are color-blind) 0 - C(,0)(0.04) (0.98)» (a) P(0 or more) - P(less than 0) -[ P(0) + P() + P() + + P(9)] [ C(00,0)(0.07) (0.97) 99 + C(00,)(0.07) (0.97) 98 + C(00, )(0.07) (0.97) 97 + C(00,)(0.07) (0.97) C(00, 4)(0.07) (0.97) 9 + C(00,)(0.07) (0.97) 94 + C(00, )(0.07) (0.97) C(00, 7)(0.07) (0.97) C(00,8)(0.7) (0.97) C(00,9)(0.07) (0.97) ]» -[ ] The probability that 0 or more will experience nausea/vomiting is about (b) P(0 or more) - P(less than 0) - [ P(0) + P() + P() + + P(9)] - [ C(00, 0)(0.7) 0 (0.99) 00 + C(00,)(0.07) (0.99) 99 + C(00, )(0.7) (0.99) 98 + C(00, )(0.07) (0.99) 97 + C(00, 4)(0.7) 4 (0.99) 9 + C(00, )(0.07) (0.99) 9 + C(00, )(0.7) ( 0.99) 94 + C(00, 7)(0.07) 7 (0.99) 9 + C(00, 8)(0.7) 8 (0.99) 9 + C(00, 9)(0.07) 9 (0.99) The probability that 0 or more will experience nausea/vomiting is about (a) Since the probability of a particular band matching is in 4 or, the probability that 4 bands match is ( ) 4 04 or chance in 04. (b) The probability that 0 bands match is (c) ( 0 )» or about chance in If 0 bands are compared, the probability that or more bands match is P(at least ) P() + P(7) + P(8) + P(9) + P(0) 4 7 æö æö æö æö C(0,) + C(0,7) è4 è4ø è4ø è4ø 8 9 æö æö æö æö + C(0,8) + C(0,9) è4 è4 è4ø è4ø 0 0 æö æö + C(0,0) è 4 è 4 Copyright 0 Pearson Education, Inc.

31 Section æö æö æö æö è4 è4 è4 è æö æö æö æö æö è4ø è4ø è4ø è4ø è4ø æö é æ 8 ö 484 æ öæ 40 7 ö æ öæ 9 ö + è4ø èø 90 + ê è4 4 ë øè ø è øè ø æ öæö ù è 4 øè 4 ø ú û æ ö æ9, , ö è4ø è ø 44, ,99»,87,0 or about chance in 4. (a) n 4, p P(at least ) - P(none) C(4, 0)(0.) (0.7)» 0.8 (b) n, p 0. P(at least ) - P(none) 0 - C(, 0)(0.) (0.7)» 0.78 (c) The assumption of independence is likely not justified because the bacteria would be present in groups of eggs from the same source. 47. n 4800, p 0.00 P(more than ) - P() - P(0) C(4800,)(0.00) (0.999) C(4800,0)(0.00) (0.999)» First, find the probability that one out of 0 vials is ineffective, given that the shipment came from company X. n 0, p 0., x 9 P() C(0,)(0.0) (0.9)» 0.4 Next, find the probability that one out of 0 vials is ineffective, given that the shipment came from company Y. n 0, p 0.0, x 9 P() C(0,)(0.0) (0.98)» 0.40 Use Bayes Theorem to find the probability that shipment came from company X. Let A be the event that the shipment came from company X and B be the event that one vial out of thirty is ineffective. PA ( ) 0. 4 PB ( )» (0.4) + (0.40)» 0.9 PBA ( )» 0.4 PBAPA ( ) ( ) PAB ( ) PB ( ) » 0.9» 0.0 The answer is a. 49. First, find the probability that one group of ten has at least 9 participants complete the study. n 0, P 0.8, P(at least 9 complete) P(9) + P(0) 9 C(0,9)(0.8) (0.) C(0,0)(0.8) (0.)» 0.78 The probability that or more drop out in one group is Thus, the probability that at least 9 participants complete the study in one of the two groups, but not in both groups, is (0.78)(0.4) + (0.4)(0.78)» The answer is e. 0. We define a success to be the event that a woman would prefer to work part-time if money were not a concern. In this situation, n 0, x number of successes, p 0.0, and - p P(at least ) P( x 0,, or ) - Px ( 0) - Px ( ) - Px ( ) C(0, 0)(0.0) (0.40) - C(0,)(0.0) (0.40) 8 - C(0, )(0.0) (0.40)» n, x 7, p P(7) C(,7)(0.8) (0.7)» 0.00 Copyright 0 Pearson Education, Inc.

32 8 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS. n, x 9, p P(9) C(,9)(0.8) (0.7)» 0.0. n, p 0.8 P(at least 9) P(9) + P(0) + P() + P() 9 0 C(,9)(0.8) (0.7) + C(,0)(0.8) (0.7) 0 + C(,)(0.8) (0.7) + C(,)(0.8) (0.7)» n, p 0.8 P(at most 9) - P(at least 0) - P(0) - P() - P() 0 - C(,0)(0.8) (0.7) - C(,)(0.8) (0.7) 0 - C(,)(0.8) (0.7)» n 0, p 0., - p 0.78 (a) (b) (c) 8 P() C(0, )(0.) (0.78)» P( or fewer) C(0,0)(0.) (0.78) 9 + C(0,)(0.) (0.78) 8 + C(0, )(0.) (0.78) 7 + C(0,)(0.) (0.78)» 0.90 If exactly do not belong to a minority, then exactly 0 - do belong to a minority, and this probability is P() C(0,)(0.) (0.78)» 0.0. (d) If or more do not belong to a minority, then at most 4 do belong to a minority, and this probability is P(at most 4) 0 0 P(at most 4) C(0,0)(0.) (0.78) 9 + C(0,)(0.) (0.78) 8 + C(0, )(0.) (0.78) 7 + C(0,)(0.) (0.78) 4 + C(0, 4)(0.) (0.78)» (a) No, the results only indicated that 84% of college students believe they need to cheat to get ahead in the world today. It says nothing about whether or not they cheat. (b) P(90 or more) P(90) + P(9) + + P(00) C(00, 90)(0.84) (0.) + C(00, 9)(0.84) (0.) C(00, 9)(0.84) (0.) + C(00, 9)(0.84) (0.) C(00, 94)(0.84) (0.) + C(00, 9)(0.84) (0.) C(00, 9)(0.84) (0.) + C(00, 97)(0.84) (0.) C(00, 98)(0.84) (0.) + C(00, 99)(0.84) (0.) C(00,00)(0.84) (0.)» The probability that 90 or more will answer affirmatively to the question is about (a) Using the binomcdf function on a graphing calculator, we find P(at least 0) - P(9 or fewer) -binomcdf (40, 0.74, 9)» (b) Using the binomcdf function on a graphing calculator, we find P(at least 0) - P(9 or fewer) -binomcdf (40, 0.8, 9)» (a) n, p 44, 0 P(at least 9) - P(at most 8) - binomcdf(, 74/440, 8)» (b) n 74, p 44, 0 P(at least 8) - P(at most 7) - binomcdf(74, 74/440, 7) -4».07 0 (c) n 74, p 74 44, 0 P(at most 4) binomcdf(7, 74/440, 4) -0» (a) Suppose the National League wins the series in four games. Then they must win all four games 4 0 and P C(4, 4)(0.) (0.) 0.0. Since Copyright 0 Pearson Education, Inc.

33 Section 8. 9 the probability that the American League wins the series in four games is equally likely, the probability the series lasts four games is (0.0) 0.. Suppose the National League wins the series in five games. Then they must win exactly three of the previous four games and P C(4, )(0.) (0.) (0.) 0.. Since the probability that the American League wins the series in five games is equally likely, the probability the series lasts five games is (0.) 0.. Suppose the National League wins the series in six games. Then they must win exactly three of the previous five games and P C(, )(0.) (0.) (0.) 0.. Since the probability that the American League wins the series in six games is equally likely, the probability the series lasts six games is (0.) 0.. Suppose the National League wins the series in seven games. Then they must win exactly three of the previous six games and P C(, )(0.) (0.) (0.) 0.. Since the Probability that the American League wins the series in seven games is equally likely, the probability the series last seven games is (0.) 0.. (b) Suppose the better team wins the series in four games. Then they must win all four 4 0 games and P C(4, 4)(0.7) (0.7)» Suppose the other team wins the series in four games. Then they must win all four games and 4 0 P C(4, 4)(0.7) (0.7)» The probability the series lasts four games is the sum of two probabilities, Suppose the better team wins the series in five games. Then they must win exactly three of the previous four games and P C(4, )(0.7) (0.7) (0.7)» Suppose the other team wins the series in five games. Then they must win exactly three of the previous four games and C(4, )(0.7) (0.7) (0.7)» 0.0. The probability the series lasts five games is the sum of the two probabilities, 0.. Suppose the better team wins the series in six games. Then they must win exactly three of the previous five games and P C(, )(0.7) (0.7) (0.7)» Suppose the other team wins the series in six Copyright 0 Pearson Education, Inc. games. Then they must win exactly three of the previous five games and P C(, )(0.7) (0.7) (0.7)» The probability the series lasts six games is the sum of the two probabilities, 0.. Suppose the better team wins the series in seven games. Then they must win exactly three of the previous six games and P C(, )(0.7) (0.7) (0.7)» 0.8. Suppose the other team wins the series in seven games. Then they must win exactly three of the previous six games and P C(, )(0.7) (0.7) (0.7)» The probability the series lasts seven games is the sum of the two probabilities, Probability Distributions; Expected Value Your Turn C(, 0) C(9, ) Px ( 0) C(, ) C(,) C(9,) 9 Px ( ) C(, ) C(, ) C(9, 0) Px ( ) C(, ) The distribution is shown in the following table: x 0 Px ( ) 9 Your Turn Let the random variable x represent the number of tails. 0 æö æö Px ( 0) C(,0) è è 8 æö æö Px ( ) C(,) è è 8 æö æö Px ( ) C(,) è èø 8

34 0 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 0 æö æö Px ( ) C(,) è èø 8 The distribution is shown n the following table: Here is the histogram. Probability P /8 /8 /8 0 Tails x 0 Px ( ) x Your Turn The expected payback is 8. Warmup Exercises W. This is a binomial experiment. The probability of a six on one roll is /. The probability of exactly n sixes in rolls is n -n æö æö C(, n) è è so for n 0,,..., these probabilities are approximately 0.409, 0.409, 0.08, 0.0, 0.00, and W. This is a binomial experiment. The probability of drawing a spade is /4. The probability of exactly n spades in 4 draws is n 4-n æö æö C(4, n) è 4 è so for n 0,,..., 4 these probabilities are approximately 0.4, 0.49, 0.09, 0.049, and æ ö æ ö æ ö æ 997 ö (-) è000 ø è000 è000 è or -$.. Your Turn 4 Let the random variable x represents the number of male engineers in the sample of. 0 Px ( 0) C(,0)(0.809) (0.9)» Px ( ) 4 C(,)(0.809) (0.9)» Px ( ) C(, )(0.809) (0.9)» Px ( ) C(,)(0.809) (0.9)» Px ( 4) C(, 4)(0.809) (0.9)» Px ( ) C(,)(0.8) (0.84)» 0.4 Ex ( )» (0)(0.00) + ()(0.008) + ()(0.040) + ()(0.9) + (4)( ) + ()(0.4) In fact the exact value of the expectation can be computed quickly using the formula Ex ( ) np. For this example, n and p so np ()(0.809) Exercises. Let x denote the number of heads observed. Then x can take on 0,,,, or 4 as values. The probabilities are as follows. 0 4 æö æö Px ( 0) C(4,0) è è æö æö 4 Px ( ) C(4,) è è 4 æö æö Px ( ) C(4,) è è 8 æö æö 4 Px ( ) C(4,) è è æö æö Px ( 4) C(4,4) è è Therefore, the probability distribution is as follows. Number of Heads 0 4 Probability Your Turn The expected number of girls in a family of a dozen æ ö children is. è Copyright 0 Pearson Education, Inc.

35 Section 8.. There are outcomes. We count the number of ways each sum can be obtained, and then divide by to get each probability. Number of Points Ways to Get This Total 4 4 Probability Let x denote the number of aces drawn. Then x can take on values 0,,, or. The probabilities are as follows. æ48 öæ47 öæ4 ö Px ( 0) C(,0)» 0.78 è øè øè 0 ø Px ( 0) + Px ( ) + Px ( ), so shade the first rectangles in the histogram.. Px³ ( ) P Use the probabilities that were calculated in Exercise, and shade the regions corresponding to x and x. x æ 4 öæ48öæ47ö Px ( ) C(,)» 0.04 è è è0 æ 4 öæ öæ48ö Px ( ) C(, )» 0.00 è è è0 æ 4 öæ öæ ö Px ( ) C(, )» è è è0 Therefore, the probability distribution is as follows. Number of Aces 0 Probability Use combinations to find the probabilities of drawing 0,, and black balls. C(,0) C(4,) P(0) C(, ) C(,) C(4,) 8 P() C(, ) C(,) C(4,0) P() C(, ) Number of Black Balls Probability Use the probabilities that were calculated in Exercise. Draw a histogram with 4 rectangles, corresponding to x 0, x, x, and x. P(at least oneace) P( x ³ ) corresponds to Px ( ) + Px ( ) + Px ( ), so shade the last rectangles. P P(at least one black ball) Use the probabilities that were calculated in Exercise 4, and shade the regions corresponding to x and x. 0. P x. Use the probabilities that were calculated in Exercise. Draw a histogram with rectangles, corresponding to x 0, x, x, x, and x 4. Px ( ) corresponds to 0 0 x Copyright 0 Pearson Education, Inc.

36 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 9. Ex ( ) (0.) + (0.4) + 4(0.) + (0.). 0. Ey ( ) 4(0.4) + (0.4) + 8(0.0) + 0(0.).9. Ez ( ) 9(0.4) + (0.) + (0.8) + 8(0.9) + (0.07) Ex ( ) 0(0.) + (0.9) + (0.) + 8(0.09) + 44(0.0).. It is possible (but not necessary) to begin by writing the histogram s data as a probability distribution, which would look as follows. x 4 P(x) Possible Results Result of toss H H Call H T Caller wins? Yes No Probability (a) Yes, this is still a fair game, since the probability of Donna matching is still. (b) If Donna calls heads, her expected gain (since she will match with probability ) is (c) (.0) $.0. If Donna calls tails, her expected gain (since she will lose with probability ) is ( -.0 ) - $.0. The expected value of x is Ex ( ) (0.) + (0.) + (0.) + 4(0.4) P() 0., P(4) 0., P() 0., P(8) 0., and P(0) 0.. Ex ( ) (0.) + 4(0.) + (0.) + 8(0.) + 0(0.).. The expected value of x is Ex ( ) (0.) + (0.) + 8(0.4) + 4(0.) + 0(0.) 8.. Notice that the probability of all values is 0.. Ex ( ) 0.( ) 0.(0) 0 7. Using the data from Example, the expected winnings for Mary are æö æö Ex ( ) è4 è4ø æö æö +. + (-.) è4 è4 0. Yes, it is still a fair game if Mary tosses and Donna calls. 9. (a) Number of Yellow Marbles 0 Probability C(, 0) C(4, ) 4 C(7, ) C(,) C(4, ) 8 C(7, ) C(, ) C(4,) C(7, ) C(, ) C(4, 0) C(7, ) Draw a histogram with four rectangles corresponding to x 0,,, and. P 0/ / 0/ / 0 x Copyright 0 Pearson Education, Inc.

37 Section 8. (b) Expected number of yellow marbles æ 4 ö æ8ö æö æ ö è è è è 4 9» (a) Use combinations to set up the probability distribution. In each case, there are rotten apples and 0 good apples to pick from. There are a total of C(, ) ways to choose any two apples. Number of Probability Simplified Rotten Apples 0 C(, 0) C(0, ) C(,) C(,) C(0,) C(,) C(, ) C(0, 0) C(.) æö æö Px ( 4) C(4,4) è è 9 x 0 4 P(x) 9 (b) æ ö æ ö æ ö Ex ( ) è9 è4 è ø æ ö æ ö è4 è9. The probability that the delegation contains no liberals and conservatives is C(,0) C(,) 0. C(, ) Similarly, use combinations to calculate the remaining probabilities for the probability distribution. (b) We therefore have æ90 ö æ00 ö æ 0 ö Ex ( ) è00 è00 è (a) Let x be the number of times is rolled. Since the probability of getting a on any single roll is, the probability of any other outcome is. Use combinations since the order of outcomes is not important. 0 4 æö æö Px ( 0) C(4,0) è è 9 æ ö æ ö Px ( ) C(4,) è è ø 4 æö æö Px ( ) C(4,) è è (a) Let x represent the number of liberals on the delegation. The probability distribution of x is as follows. x 0 P(x) 0 7 The expected value is 0 0 æ 0 ö æ 7 ö Ex ( ) 0 + è è æ 0 ö æ 0 ö + + è è ø». liberals. (b) Let y represent the number of conservatives on the committee. The probability distribution of y is as follows. y 0 P(y) æö æö Px ( ) C(4,) è è 4 Copyright 0 Pearson Education, Inc.

38 4 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS The expected value is æ 0 ö æ 0 ö Ey ( ) 0 + è è æ 7 ö æ 0 ö + + è è ø 70 8».4 conservatives.. Set up the probability distribution. Number of 0 Women Probability Simplified C(,0) C(,) C(8,) 4 C(,) C(,) C(8,) 8 æ ö æö æ ö Ex ( ) è4 è8 è C(,) C(,0) C(8,) 4. Let x represent the number of junior members on the committee. Use combinations to find the probability of 0,,,, and 4 junior members. C(0, 4) 7, 40; C(0, 0) C(0, 4) 484; C(0, ) C(0, ), 400; C(0, ) C(0, ) 80; C(0, ) C(0, ) 400; C(0, 4) C(0, 0) 0; The probability distribution of x is as follows. P(x) x æ ö æ 70 ö æ90 ö æ 0 ö Ex ( ) è87 è87 ø è09 ø è87 ø æ ö + 4 è 4. The expected number of junior members is.. Set up the probability distribution as in Exercise 0. Number of Women Probability Simplified 0 C(,0) C(9,) C(,) C(,) C(9,) C(,) C(,) C(9,0) C(,) æ 74 ö æ 07 ö æ 78 ö Ex ( ) è è è. The probability of drawing diamonds is C(, ) 78, C(, ) and the probability of not drawing diamonds is Let x represent your net winnings. Then the expected value of the game is æ 78 ö æ48 ö Ex ( ) 4. + (-0.) è è 7 -»-$0. or -. No, the game is not fair since your expected winnings are not zero. 9. (a) First list the possible sums,,, 7, 8, and 9, and find the probabilities for each. The total possible number of results are 4. There are two ways to draw a sum of ( then, and then ). The probability of is. There are two ways to draw a sum of ( then 4, and 4 then ). The probability of is. There are four ways to draw a sum of 7 ( then, then 4, 4 then, and then ). The probability of 7 is 4. There are two ways to draw a sum of 8 ( then, and then ). The probability of 8 is. There are two ways to draw a sum of 9 (4 then, and then 4).The probability of 9 is. The distribution is as follows. Sum (b) Probability P x Copyright 0 Pearson Education, Inc.

39 Section 8. (c) (d) The probability that the sum is even is + Thus the odds are to.. Ex ( ) () + () + (7) + (8) + (9) 7 0. The expected value is Ex ( ) 0(0.0) + (0.0) + (0.) + (0.) + 4(0.) + (0.) + (0.0).4complaints per day.. We first compute the amount of money the company can expect to pay out for each kind of policy. The sum of these amounts will be the total amount the company can expect to pay out. For a single $00,000 policy, we have the following probability distribution. Pay Don t Pay Outcome $00,000 $00,000 Probability E (payoff ) 00, 000(0.00) + 0(0.9998) $0 For all 00 such policies, the company can expect to pay out 00(0) $, 000. For a single $0,000 policy, E (payoff ) 0, 000(0.00) + 0(0.9998) $0. For all 00 such policies, the company can expect to pay out 00(0) $0, 000. Similarly, for all 000 policies of $0,000, the company can expect to pay out 000() $, 000. Thus, the total amount the company can expect to pay out is $,000 + $0,000 + $,000 $4,000.. Let x represent the benefit amount. x P(x) x P(x) Ex ( ) 4000(0.4) + 000(0.4) + 000(0.44) + 000(0.084) + 0(0.9) 94.4 The answer is e.. (a) Expected number of good nuts in 0 blow outs is 4. Ex ( ) 0(0.0) 0. (b) Since 80% of the good nuts are good, 0% are bad. Expected number of bad nuts in 0 good nuts is Account Number Ex ( ) 0(0.0) 0. Expected Value Exist. Vol. + Exp. Value Class $ 00 $7,000 C $40,000 C $ 000 $,000 C 4 $ 000 $,000 B $,000 $0,000 C $0,000 $0,000 A 7 $,000 $4,000 B. The tour operator earns $00 if or more tourists do not show up. The tour operator earns $90 if all tourists show up. The probability that all tourists show up is (0.98)» 0.4. The expected revenue is 00(0.47) + 90(0.4) The answer is e.. Let x represent the number of offspring. We have the following probability distribution. x 0 4 P(x) Ex ( ) 0(0.9) + (0.) + (0.8) + (0.) + 4(0.4). Copyright 0 Pearson Education, Inc.

40 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 7. (a) Expected cost of Amoxicillin: Ex ( ) 0.7($9.0) + 0.($9.) $8. Expected cost of Cefaclor: Ex ( ) 0.90($9.) + 0.0($0.00) $7.84 (b) Amoxicillin should be used to minimize total expected cost. 8. Calculate the probability and payment for the number of days of hospitalization. X P(x) Payment 4 Expected payment is $00 4 $00 $00 $ $0 æö æ 4 ö æö æ ö è èø èø èø æ ö + 0 è». The answer is d. 9. Ex ( ) 0(0.74) 8 We would expect 8 low-birth-weight babies to graduate from high school. 40. Expected number of a group of 00 college students that say they need to cheat is Ex ( ) 00(0.84) (a) Using binomial probability, n 48, x 0, p P(0) C(48,0)(0.097) (0.904)» (b) Using combinations, the probability is C(74, 48) -4» C(8, 48) (d) Using binomial probability, n, p 0.. P(at least ) - C(,0)(0.) (0.9) - C(,)(0.) (0.9)» (a) Let x represent the amount of damage in millions of dollars. For seedling, the expected value is Ex ( ) 0.08(.8) + 0.4(9.) + 0.9(00) + 0.(4.7) + 0.7(.)» $94.0 million. For not seedling, the expected value is Ex ( ) 0.04(.8) + 0.0(9.) (00) + 0.0(4.7) (.)» $.0million. (b) Seed, since the total expected damage is less with that option. 4. (a) The six possibilities and their scores are as follows: æ4öæ4ö P(neither nor ) è è score: 0 æöæ4ö P(exactly one ) è è score: 0 æöæ4ö P(exactly one ) è è score: 00 æöæö P(one, one ) è è ø score: 0 æöæö P(two s) è è score: 00 æöæö P(two s) è è score: 00 The expected score is é æ öæ öù é æ öæ öù é æ öæ öù êë è øè øúû êë è øè øúû êë è øè øúû (0) + (00) + (0) éæ öæ öù éæ öæ öù + (00) + (00) 0 ê è è ú êè è øú ë û ë û (c) Using binomial probability, n, x, p P(0) C(,)(0.) (0.9) + (0.). 0 Copyright 0 Pearson Education, Inc.

41 Section 8. 7 (b) The ten possibilities and their scores are æ4 ö P(neither nor ) è score: 0 æöæ4ö P(exactly one ) è èø score: 0 æöæ4ö P(exactly one ) è èø score: 00 æö æ4ö P(one, one ) è è score: 0 æö æ4ö P(two s) è è score: 00 æö æ4ö P(two s) è è score: 00 æö æö P(two s, one ) è èø score: 00 æö æö P(one, two s) è è ø score: 0 æ ö P(three s) è score: 0 æ ö P(three s) è score: 00 The expected score is é ù é ù (0) æ öæ ö æ öæ ö + (00) ê è øè ø è øè ø ë úû êë úû é 4 ù é 4 ù + (0) æ ö æ ö æ ö æ ö + (00) ê è ø è ø è ø è ø ë úû êë úû é æö æ 4öù é + (00) ù + (00) æ ö æ ö è è ê è ø è ø ë úû êë úû é ù é ù é ù (0) æ ö æ ö æ ö æ ö è è è èø êë úû êë úû êë úû (c) We can subtract the last two terms from the expected value sum in (b) and in their place add the term ( ) æ ö è which yields an expected value of æö æö » 8.8 è èø 44. (a) We define a success to be the event that a letter was delivered the next day. In this situation, n 0; x 0,,,,..., 0; p 0.8, and - p 0.7. Number of Letters Delivered the Probability Next Day C (0,0)(0.8) (0.7)» C (0,)(0.8) (0.7)» C (0, )(0.8) (0.7)» C (0, )(0.8) (0.7)» C (0, 4)(0.8) (0.7)» C (0, )(0.8) (0.7)» C (0, )(0.8) (0.7)» C (0, 7)(0.8) (0.7)» C (0,8)(0.8) (0.7)» C (0, 9)(0.8) (0.7)» C (0,0)(0.8) (0.7)» 0. (b) P(4or fewer letters would bedelivered) Px ( 0) + Px ( ) + Px ( ) + Px ( ) + Px ( 4)» (d) Expected number of letters delivered next day» 0(0.0000) + (0.0000) + (0.0000) + (0.000) + 4(0.004) + (0.04) + (0.07) + 7(0.00) + 8(0.99) + 9(0.78) + 0(0.)» Below is the probability distribution of x, which stands for the person s payback. P(x) x $98 $78 - $ The expected value of the person s winnings is Ex ( ) 98(0.00) + 78(0.00) + (-)(0.994)»-$0.7 or -7. Since the expected value of the payback is not 0, this is not a fair game. Copyright 0 Pearson Education, Inc.

42 8 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 4. Reduce each price by the 0 cost of the raffle ticket, and multiply by the corresponding probability. æ ö æ ö Ex ( ) è0,000 è0,000 æ 0 ö æ 9977 ö (-0.0) è0,000 è0, $0. or - 0,000 No, this is not a fair game. In a fair game the expected value is There are possible outcomes for each suit. 4 That would make 8, total possible outcomes. In one case, you win $000 (minus the $ cost to play the game). In the other 8,0, cases, you lose your dollar. æ ö æ8, 0 ö Ex ( ) (-) è8, è 8, The probability of getting exactly of the 4 selections correct and winning this game is æ ö æ ö» è è C(4, ) The probability of losing is If you win, your payback is $99. Otherwise, you lose $ (win- $). If x represents your payback, then the expected value is Ex ( ) 99(0.008) + (-)(0.9989) »-$0. or -. If an even number comes up, you win $. Otherwise, you lose $ (win- $). If x represents your payback, then the expected value is æ8 ö æ9 ö Ex ( ) + (-) è7 è7 ø -»-$0.07 or You have one chance in a thousand of winning $00 on a $ bet for a net return of $499. In the 999 other outcomes, you lose your dollar. æ ö æ 999 ö Ex ( ) (-) è000 è In this form of the game Keno, 0 P (your number comes up) 80 4 and 0 P (your number doesn't come up) If your number comes up, you win $.0. Otherwise, you lose $ (win- $). If x represents your payback, then the expected value is æö æö Ex ( ) è4 è4 -$0.0 or -0.. Let x represent the payback. The probability distribution is as follows. x P(x) 00,000,000, There are possible outcomes. In 8 cases you win a dollar and in 0 you lose a dollar; hence, æ8 ö æ0 ö Ex ( ) + (-) è8 è8 -, or about ,000 0,000 0 The expected value is,000,000,000,000,999,99,000, In this form of roulette, 8 9 P(even) and P(noneven). 7 7 Copyright 0 Pearson Education, Inc.

43 Section 8. 9 æ ö æ ö Ex ( ) 00, , 000 è,000,000 è,000,000 æ ö æ,999,99 ö + 0, è,000,000 è,000, $ Since the expected payback is 0, if entering the context costs 00, then it would be worth it to enter. The expected net return is - $ At any one restaurant, your expected winnings are æ ö æ ö Ex ( ) 00,000 +,000 è7, 40,00 è9, 00, ø æ ö æ ö è7,40, 0 è,8,0 æ ö æ ö æ ö è88, 44 è70 ø è88 ø Going to restaurants gives you expected earnings of ( ) 0.. Since you spent $, you lose 87.8 on the average, so your expected value is (a) The possible scores are 0,,, 4,,. Each score has a probability of. æö æö æö Ex ( ) è èø è æö æö æö è è è 0 (0) (b) The possible scores are 0 which has a probability of, 4 which has a probability of, which has a probability of, which has a probability of, 7 which has a probability of 4, 8 which has a probability of, 9 which has a probability of 4, 0 which has a probability of, which has a probability of, which has a probability of. æ ö æ ö æ ö æ ö æ 4 ö Ex ( ) è è è ø è ø è ø æ ö æ 4 ö æ ö æ ö æ ö è è ø è ø è ø è (c) If a single die does not result in a score of zero, the possible scores are,, 4,, with each of these having a probability of. æö æö æö æö æö Ex ( ) èø èø èø èø è (0) 4 Thus, if a player rolls n dice the expected average score is n E( x) n 4 4 n. (d) If a player rolls n dice, a nonzero score will occur whenever each die rolls a number other than. For each die there are possibilities so the possible scoring ways for n n dice is. When rolling one die there are possibilities so the possible outcomes for n dice is. n The probability of rolling a scoring set of dice is n ; thus the expected n value of the player s score when rolling n (4) dice is ( ) n n Ex.. (a) Expected value of a two-point conversion: Ex ( ) (0.478) 0.9 n Expected value of an extra-point kick: Ex ( ) (0.99) 0.99 (b) Since the expected value of an extra-point kick is greater than the expected value of a two-point conversion, the extra-point kick will maximize the number of points scored over the long run. Copyright 0 Pearson Education, Inc.

44 0 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS 7. Let x represent the number of hits. Since p 0., - p P(0) C(4,0)(0.) (0.9) 0.00 P() C(4,)(0.) (0.9) 0.94 P() C(4, )(0.) (0.9) 0.94 P() C(4,)(0.) (0.9) P(4) C(4,4)(0.) (0.9) 0.00 The distribution is shown in the following table. x 0 4 P(x) The expected number of hits is np (4)(0.).4. Chapter 8 Review Exercises. True. True. True 4. True. False: The probability of at least two occurrences is the probability of two or more occurrences.. True 7. True 8. False: Binomial probability applies to trials with exactly two outcomes. 9. True 0. False: For example, the random variable that assigns 0 to a head and to a tail has expected value / for a fair coin.. True. False: The expected value of a fair game is 0.. shuttle vans can line up at the airport in P (, )! 70 different ways. 4. Since order makes a difference, use permutations.!! P (,) ( - )!! 4 0 There are 0 variations in first-, second- and third-place finishes.. oranges can be taken from a bag of in! 0 C(, ) 0 9!! different ways.. Since the order of selection is not important, use combinations. 0! C (0, 4) 0!4! 7. (a) The sample will include of the rotten oranges and of the 0 good oranges. Using the multiplication principle, this can be done in C(,) C (0,) 4 90 ways. (b) The sample will include both of the rotten oranges and of the 0 good oranges. This can be done in (c) C(,) C (0,) 0 0 ways. The sample will include 0 of the rotten oranges and of the 0 good oranges. This can be done in C(, 0) C (0, ) 0 0 ways. (d) If the sample contains at most rotten oranges, it must contain 0,, or rotten oranges. Adding the results from parts (a), (b), and (c), this can be done in ways. 8. (a) Exactly males means males out of and female out of 4: C(,) C (4,) (b) No males means all 4 females must be selected, and there is way of doing this. (c) At least males means, or 4 males: C(, ) C(4,) + C(,) C(4,) + C(,4) C(4,0) (a) P (,)! 0 (b) P (4,4) 4! 4 0. (a) There are! ways to arrange the landscapes,! ways to arrange the puppies, and Copyright 0 Pearson Education, Inc.

45 Chapter 8 Review choices whether landscape or puppies come first. Thus, the pictures can be arranged in!! 4 different ways. (b) The pictures must be arranged puppy, landscape, puppy, landscape, puppy. Arrange the puppies in! or ways. Arrange the landscapes in! or ways. In this scheme, the pictures can be arranged in different ways.. There are C(, ) ways to choose the balls and C(4, ) ways to get all black balls. Thus, C(4,) 4 P(all black) C(, ) 8» It is impossible to draw blue balls, since there are only blue balls in the basket; hence, P (all blue balls) 0.. (a) The order within each list is not important. Use combinations and the multiplication principle. The choice of three items from column A can be made in C(8, ) ways, and the choice of two from column B can be made in C(, ) ways. Thus, the number of possible dinners is C(8, ) C (, ) 840. (b) There are C(8, 0) + C(8,) + C(8, ) + C(8,) ways to pick up to items from column A. Likewise, there are C(,0) + C(,) + C(, ) ways to pick up to items from column B. We use the multiplication principle to obtain [ C(8,0) + C(8,) + C(8,) + C(8,) ] [ C(,0) + C(,) + C(,) ] ( )( + + ) 9() 04. Since we are assuming that the diner will order at least one item, subtract to exclude the dinner that would contain no items. Thus, the number of possible dinners is 04.. (a) There are different groups of representatives possible. (b) is the number of groups with representatives. For representatives, the number of groups is For representative, the number of groups is The total number of these groups is groups. Copyright 0 Pearson Education, Inc. 7. There are C(4, ) ways to get black balls and C(7, ) ways to get green ball. Thus, C(4,) C(7,) P( black and green) C(,) (.7) 4» P(exactly black balls) C(4,) C(9,) 4 7» C(, ) There are C(,) ways to get blue ball and C(, ) ways to get nonblue balls. Thus, C(,) C(,) P(exactlyblue) C(,) 0» P( green balls and blue ball) C(7, ) C(,) 4» 0.49 C(,) 8 4. This is a Bernoulli trial problem with P(success) P(girl). n, p, and x. Here, æö æö P(exactlygirls) C(,) è è ø 0» 0. 4

46 Chapter 8 COUNTING PRINCIPLES; FURTHER PROBABILITY TOPICS. Let x represent the number of girls. We have n, x, p, and - p, so 0 æö æö P(all girls) C(, ) è è» P(at least 4 girls) P(4girls) + P(girls) + P(girls) æö æö æö æö C(,4) + C(,) è è èø èø 0 æö æö + C(,) è è» Let x represent the number of boys, and then p and - p. We have P(nomore than boys) Px ( ) Px ( 0) + Px ( ) + Px ( ) 0 æö æö æö æö C(,0) + C(,) è è è è 4 æö æö + C(,) è è» P(both red). C(,)» 0.4 C(, ) 0 C(, ) P(spades) C(, ) 78 7» P(at leastcard is a spade) - P(neither is a spade) C(9, ) C(, ) 8» P(exactly face card) C(,) C(40,) C(, )» There are face cards and 40 nonface cards in an ordinary deck. P(at least face card) P( face card) + P( face cards) C(,) C(40,) C(, ) + C(, ) C(, ) » P(at most queen) P(0 queens) + P( queen) C(48,) C(4,) C(48,) + C(, ) C(, ) » This is a Bernoulli trial problem. (a) P(success) P(head). Hence, n and p. Number Probability of Heads (b) 0 C (, 0) 0. 0 ( ) ( ) C (,) 0.7 ( ) ( ) C (, ) 0.7 ( ) ( ) 0 C (, ) 0. ( ) ( ) Copyright 0 Pearson Education, Inc.