Solutions - Problems in Probability (Student Version) Section 1 Events, Sample Spaces and Probability 1. If three coins are flipped, the outcomes are HTT,HTH,HHT,HHH,TTT,TTH,THT,THH. There are eight outcomes. We can also count these outcomes as follows. An outcomes has the form (x, y, z), where each of x, y and z is an H or a T, independently. There are two possibilities for x, and for each, two for y for four possibilities from the first two flips. There are also two possibilities for z, with each of these four possibilities from the first two flips, making 8 outcomes in all. (a) There are three outcomes with exactly two heads, so Pr(exactly two heads) = 3 8. (b) For at least two heads, we could have exactly two heads (three outcomes) or all three heads (one outcome), for four ways this can happen. Then Pr(at least two heads) = 4 8 = 1 2. 3. Three dice are rolled. With six possibilities independently for each die (the number that comes up on one die does not influence what comes up on another), there are 6 6 6 = 216 outcomes. (a) The dice can come up the same in six ways, namely: Then (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6). Pr(all dice come up the same) = 6 216. (b) The outcomes with the dice totalling 15 are (6, 6, 3), (6, 5, 4), (6, 4, 5), (6, 3, 6), (5, 6, 4), (5, 5, 3), (5, 4, 6), (4, 6, 5), (4, 5, 6), (3, 6, 6). These have been displayed in a way to suggest the systematic way they were listed. In counting objects by enumerating them, it is helpful to have a scheme in mind to avoid duplications or omissions. In this case, we 1
started with outcomes having 6 on the first die, then those having 5, and so on. There are 10 outcomes with the dice totaling 15. Then Pr(the dice total )15 = 10 216. (c) Suppose the first die comes up 1 and the third comes up 4. This leaves six outcomes, since the second die can come up 1, 2, 3, 4, 5 or 6. Therefore Pr(first die 1, second die 4) = 6 216 (d) Suppose the first two dice come up even. There are now three ways the first die can come up, and, independently, three ways the second die can come up, and, independent of these two tosses, six ways the third die can come up. This is a total of 3 3 6, or 54 outcomes. Then Pr(first two dice come up even) = 54 216 = 1 4. 5. Flip two coins and then roll a die. We can record an outcome as (x, y, z), with x and y can independently be H or T and z can be any integer 1, 2,, 6. There are 2 possibilities for each coin and 6 for the dice, yielding 2(2)(6) = 24 outcomes. (a) Suppose one head, one tail and an even number come up. If the head comes up first, then the other coin is T and there are three possibilities for the die (2, 4 or 6). If the tail comes up first, there are also three possibilities for the two coins and the die. There are therefore 6 outcomes with one head, one tail and an even number on the die. Then Pr(one H, one T, and an even number) = 6 24 = 1 4. (b) There are two outcomes with both coins heads and the die a1or4. These outcomes are (H, H, 1) and (H, H, 4). Then Pr(both heads, 1 or 4 on the die) = 2 24 = 1 12. (c) Suppose at least one tail comes up, with a 4, 5 or 6 on the die. Now the outcomes all have the form (T,H,x), (H, T, x) or(t,t,x) in which x can be 4, 5 or 6. This yields 9 outcomes. Therefore Pr(at least one tail, and 4, 5 or 6) = 9 24 = 3 8. 2
Section 2 Four Counting Principles 1. The fact that these are letters of the alphabet that we are arranging in order is irrelevant. The issue is that there are nine distinct objects. The number of arrangements is 9!, which is 362, 880. 3. Since any of the nine integers can be used in any of the nine places of the ID number, there are 9 ways the first digit can be chosen, 9 ways the second digit can be chosen, and so on. The total number of codes is 9 9, or 3.87420489(10) 8. 5. These 7 symbols have 7! = 5, 040 permutations or arrangements. If a is fixed as the first symbol, then there are six symbols to choose in any order for the other six places, and the number of choices is 6! = 720. If a is fixed as the first symbol, and g as the fifth, then we have five symbols left to choose in any order for the remaining five places. The number of ways of doing this is 5! = 120. 7. There are 12! = 479, 001, 600 outcomes. 9. We want to pick 7 objects from 25, taking order into account. The number of ways to do this is 25P 7 = 25! = (19)(20)(21)(22)(23)(24)(25) = 2, 422, 728, 000. 18! 11. Because order is important, the number of possibilities is 12. (a) The number of choices is 22P 6 = 22! =53, 721, 360. 16! 20P 3 = 20! =6, 840. 17! (b) If the list begins with 4, there are only two numbers to choose from the remaining nineteen numbers. There are 19P 2 = 19! = (18)(19) = 342 17! ways to do this. This result does not depend on which number is fixed in the first position. The percentage of choices beginning with 4 is 100(342/6, 840), or 5 percent. This is reasonable from a common sense point of view, since, with 20 number to choose from, we would expect 5 percent to begin with any particular one of the numbers. 3
(c) This question is really the same as the question in (b), since it does not matter which number is fixed. The answer is that 5 percent of the choices ends in 9. (d) If two places are fixed at 3 and 15, then there are eighteen numbers left, from which we want to choose one. There are 18 such choices. 13. Without order (and, we assume, without replacement), the number of ten card hands is 52C 10 = 52! =15, 820, 024, 220. 10!42! 15. The number of combinations is 20C 4 = 20! =4, 845. 4!16! 17. The number of outcomes of flipping five coins is 2 5 = 32. (a) The number of ways of getting exactly two heads from the five flips is 5 C 2 =5!/(2!3!) = 10. The probability of getting exactly two heads (or exactly two tails) is Pr(exactly two heads) = 10 32 = 5 16. (b) We get at least two heads if we get exactly two, or exactly three, or exactly four, or exactly five heads. The sum of the number of ways of doing each of these is Therefore 5C 2 + 5 C 3 + 5 C 4 + 5 C 5 =10+10+5+1=26. Pr(at least two heads) = 26 32 = 13 16. 19. The number of ways of drawing two cards out of 52 cards, without regard to order, is 52 C 2 =1, 326. (a) We get two kings if we happen to get two of the four kings, and there are 4 C 2 = 6 ways of doing this. Then Pr(two kings are drawn) = 6 1326 = 1 221. (b) The aces and face cards constitute 16 of the 52 cards. If none of the two cards is drawn from these sixteen cards, then the two cards are drawn from the remaining 36 cards. Disregarding order, there are 36 C 2 = 630 ways to do this. Therefore Pr(no ace or face card) = 630 1326 = 105 221. 4
21. The number of ways of choosing, without order, three of the eight bowling balls is 8 C 3 = 56. (a) For none of the balls to be defective, they had to come from the six nondefective ones. There are 6 C 3 = 20 ways to do this. Then Pr(none defective) = 20 56 = 5 14. (b) There are two ways to take one defective ball. The other two would have to be taken from the six nondefective ones, which can be done in 6C 2 = 15 ways. Then Pr(exactly one defective ball) = 30 56 = 15 28. (c) In choosing three bowling balls, there are 3 ways of picking the two defective ones. Then there are six ways of choosing the third ball as nondefective. Therefore Pr(both defective balls are chosen) = 6(3) 56 = 9 28. 23. Taking order into account, there are 20P 5 = 20! = (16)(17)(18)(19)(20) = 1, 860, 480 15! ways to choose 5 of the balls. (a) There is only one way to choose the balls numbered 1, 2, 3, 4, 5 in this order. the probability is Pr(select 1, 2, 3, 4, 5 in this order) = 1 1860480. (b) We want the probability that ball number 3 was drawn (somewhere in the five drawings). We therefore need the number of ordered choices of five of the twenty numbers, that include the number 3. We can think of this as choosing, in order, four of the nineteen numbers 1, 2, 4, 5,, 18, 19, 20, and then inserting 3 in any of the positions from the first through fifth numbers of the drawing. This will result in all ordered sequences of length five from the twenty numbers, and containing the number 3 in some position. There are therefore 5( 19 P 4 ) = 465, 120 such sequences. Therefore Pr(selecting a 3) = 465120 1860480 = 1 4. (c) We must count all the drawings (sequences) that contain at least one even number. This could mean the sequence contains one, two, three, 5
four or five even numbers. This is a complicated counting problem if approached directly. It is easier to count the sequences that have no even number, hence are formed just from the ten even integers from 1 through 20. If this number is N 0, then the number X of sequences having an even number is X =1, 860, 480 N 0. Now N 0 is easy to compute, since this is the number of ordered five-term sequences of the ten odd numbers. Thus N 0 = 10 P 5 =30, 240. Then X = 1860480-30240 = 1830240. Then Pr(an even number was drawn) = 1830240 1860480. This is approximately 0.984, so it is very likely that an even numbered ball was drawn. 25. The number of five-card hands (disregarding the order of the deal) is 52C 5 =2, 598, 960. (a) The number of hands containing exactly one jack and exactly one king is 4(4)( 44 C 3 ) = 211, 904. This is because there are four ways of getting one jack, four ways of getting one king, and then we choose (without order) three cards from the remaining 44 cards. Thus Pr(exactly one jack and exactly one king) = 211904 2598960. This is approximately 0.082, so this event is quite likely. (b) The hand will contain at least two aces if it has exactly two aces, exactly three aces, or exactly four aces. The number of such hands is ( 4 C 2 )( 48 C 3 )+( 4 C 3 )( 48 C 2 )+( 4 C 4 )( 48 C 1 ) = 108, 336. For the first term, choose two aces out of four, then three cards from the remaining forty-eight cards, and similarly for the other two terms for drawing three aces or drawing four aces. Then Pr(draw at least two aces) = 108336 2598960. This is approximately 0.042. As we might expect, a hand with at least two aces is not very likely. 6
Section 3 Complementary Events 1. There are many ways seven dice can come up with at least two fours (call this event E). We can count these, but it may be easier to look at the complementary event E C, which is that fewer than two dice come up 4. This is the event that exactly one die comes up 4, or none of them do. If no 4 comes up, then each of the seven dice has five possible numbers showing, for 5 7 possibilities. If exactly one die comes up with a four, then the other dice have five possible numbers showing, and this can occur in 7(5 6 ) ways. This means that E C has 5 7 + 7(5 6 ) = 187, 500 outcomes. Then, since the total number of outcomes of seven rolls is 6 7 = 279, 939, we have Pr(E C )= 187500 279936. The probability we are interested in is This is approximately 0.330. Pr(E) = Pr(at least one 4) = 1 187500 279936. 3. Draw 5 cards from the deck. Without regard to the order of the draw, there are 52 C 5 =2, 598, 960 hands that can occur. Consider the event E that at least one card is a face card, or is numbered 4 or higher. E will take some work to enumerate. Thus look at the smaller complementary event E C, which is the event that the hand has no ace, jack, king or queen, and every card numbers 2 or 3. This leaves just two s and three s to form the hand. With these eight cards, the number of possible hands is 8 C 5 = 56. Then Pr(E C 56 )= 2598960. Then which is approximately 0.999. Pr(E) =1 56 2598960, 5. Let E be the event that at least one of the four numbers selected from 1, 2,, 55 is greater than 4. To compute Pr(E) we must count the outcomes in E, and this is tedious. Look at the apparently simpler event E C, which is the event that all four numbers chosen are less than or equal to 4, hence must be chosen from the numbers 1, 2, 3, 4. Since we are choosing four numbers, there is only one way to do this (without regard to order) from these four numbers. Since 7
the total number of ways of choosing four numbers from fifty-five numbers is 55 C 4, then Pr(E C )= 1 1 = 55C 4 341055. Then 1 Pr(E) =1 341055, which is nearly 1, as we might expect intuitively. Section 4 Conditional Probability 1. (a) The experiment has four outcomes. Let E be the event that the first coin comes up heads. Then E = {HH,HT}, and Pr(E) = 1 2. (b) Let U be the event that at least one coin came up heads. Then U = {HH,HT,TH} and E U = {HH,HT}. The conditional probability of E knowing U is number of outcomes ine U Pr(E U) = number of outcomes in U = 2 3. Alternatively, we could compute Pr(E U) = Pr(E U) Pr(U) = 2/4 3/4 = 2 3. 3. Flip four coins. There are 2 4 = 16 outcomes. (a) Let E be the event that at least three come up tails. Then E has five outcomes, namely TTTH,TTHT,THTT,HTTT,TTTT. Then Pr(E) = 5 16. (b) We want the probability that at least three come up tails, having seen two come up tails. This means we know that there were at least two tails. Let U be the event that at leat two tails come up. Then U consists of the outcomes HHTT,HTHT,THHT,THTH,TTHH, HTTH,TTTH,TTHT,THTT,HTTT,TTTT. Then Pr(U) =11/16. Further, E U = E, so Pr(E U) = Pr(E U) Pr(U) = Pr(E) Pr(U) = 5 16 = 5 11. 8
5. Roll dice four times. First we want the probability that the dice total exactly 19. The number of outcomes of the four rolls is 6 4 =1, 296. Now we want the number of outcomes in the event E that the dice total 19. Here are the ways the dice can total 19: 6, 6, 6, 1 ( four ways), 6, 6, 5, 2 (twelve ways) 6, 6, 4, 3 ( twelve ways), 6, 5, 5, 3 ( twelve ways) 6, 5, 4, 4 ( twelve ways), 5, 5, 5, 4 ( four ways). E has 56 outcomes in it, and Pr(E) = 56 1296, which is about 0.043. Now suppose we want the probability that the dice total exactly 19, given that one die came up 1 (let this event be U). Consistent with this, we could have one, two, three or four dice come up 1. We need to count the number of outcomes in U. But U consists of all outcomes except those in which no die came up 1. There are 5 4 = 625 such outcomes. Therefore U has 6 5 5 4 = 1296 625 = 671 outcomes. Further E U consists of outcomes totalling 19 and having at least one 1, and there are only four such outcomes. Therefore Pr(E U) = 4 671, which is approximately 0.059. 7. Deal a five card hand without regard to order. We want the probability that the hand has four aces, knowing that a four of spades was dealt. Let E be the event that the hand has four aces, and U the event that a four of spades was dealt. Now E U consists of all outcomes in which there are four aces and a four of spades. Disregarding order, there is one five card hand having four aces and a four of spades, hence one outcome in E U. We need to count the outcomes in U. If one card is a four of spades, the other four can be any four of the remaining fifty-one cards. There are therefore 51C 4 = 249, 900 unordered hands having the four of spades. Then which is very nearly zero. Pr(E U) = 1 249, 900, 9
9. Toss four dice. We want the probability that all four come up an odd number. This probability is Pr(all odd in four tosses) = 34 6 4 = 1 16. Now let E be the event that all four dice come up odd and let U be the event that one die came up 1 and another, 5. We want Pr(E U). The number of outcomes in U is 2( 4 C 2 )(6)(6), or 432. Next, E U consists of all outcomes in which it is know that one die comes up 1 and one comes up 5, and all four come up odd. this means we roll 1, 5, odd, and odd, and there are 2( 4 C 2 )(3)(3) = 108 ways to do this. Then Pr(E U) = 108 432 = 1 4. Finally, suppose we know that the second die came up 6. Now the outcome that all four dice came up odd is not in the conditional sample space, and may be thought of as having probability 0. Section 5 Independent Events 1. Flip four coins. E is the event that exactly one coin comes up heads, and U is the event that at least three come up tails. Now, E U consists of the outcomes in which one coin comes up heads and the others are all tails. This can happen in four ways, so Pr(E U) = 4 2 4 = 1 4. But Pr(E) = and Pr(U) =5/16. Since Pr(E U) Pr(E)Pr(U), then E and U are not independent. This is intuitively apparent, since the knowledge that exactly one coin comes up heads influences the probability that at least three come up tails, by removing one possible way this can happen. 3. Two dice are rolled and E is the event that they total more than 11. U is the event that at least one die comes up even. E can happen only one way (they both come up 6), so Pr(E) = 1 6 2 = 1 36. 10
And U consists of all outcomes with exactly one die coming up even (eighteen ways this can happen) and all outcomes with both even (nine ways). Therefore Pr(U) = 27 36 = 3 4. Finally, E U consists of all outcomes with at least one even, and totalling more than 11. This can happen in one way (both dice 6), so Since these events are not independent. Pr(E U) = 1 36. Pr(E U) Pr(E)Pr(U), 5. Flip two coins and roll two dice. E is the event that at least one head comes up. U is the event that at least one 6 comes up. E U is the event that at least one coin comes up heads and at least one die comes up 6. The number of outcomes in E is (3)(36), which is 108. Then Pr(E) = 108 2 2 6 2 = 3 4. The number of outcomes in U is 4(11), so Pr(U) = 44 4(36) = 11 36. Finally, E U consists of all outcomes with at least one head and at least one 6. There are 3(11) = 33 outcomes in this event. Therefore Pr(E U) = 33 4(36) = 11 48. Now observe that Pr(E)Pr(U) = 3 11 4 36 = 11 = Pr(E U), 48 so E and U are independent. 7. Deal two cards from a 52 card deck, without replacement. E is the event that the first card drawn was a jack of diamonds. U is the event that the second card was a club or spade. E U is the event that the first card was a jack of diamonds and the second card a club or spade. E can happen in 51 ways, since the deal is without replacement. then Pr(E) = 51 52(51) = 1 52. 11
We must count the outcomes in U. If the first card drawn was a club or spade, there are 26 choices for this card, and then 25 for the second, for (26)(25) outcomes. If the first card drawn was not a club or spade, then there are 26 choices for the first card and 26 for the second as well for 26 2 outcomes. Then U has 26(25 + 26), or (26)(51) outcomes. Then Pr(U) = 26(51) 52)(51) = 1 2. Finally, E U has 26 outcomes in it, so Pr(E U) = 26 52(51) = 1 102. Now compute Pr(E)Pr(U) = 1 1 52 2 = 1 104, so E and U are not independent. 9. There are 2 4 = 16 outcomes of four flips. The following have at least two heads: HHHH,HHHT,HHTH,HTHH,THHH,HHTT, HTHT,HTTH,TTHH,THTH,THHT. The probability of a head is 0.4, so the probability of a tail must be 0.6. Then Pr(at least two heads) = (0.4) 4 + 3(0.4) 3 (0.6) + 6(0.4) 2 (0.6) 2 =0.4864. The probability of getting exactly two heads is which is approximately 0.3456. Pr(exactly two heads) = 6(0.4) 2 (0.6) 2, 11. Let r be the probability of drawing a red, g the probability of a green, and b the probability of a blue marble. Since any drawn marble must be blue, green or red, then r + b + g =1. But b =2g =3r, so so r +3r + 2 3 r = 11 2 r =1 r = 2 11,g = 3 2 2 11 = 3 11, and b = 6 11. 12
Now we know the probability of picking any particular color of a marble out of the jar. Imagine picking three marbles. We want the probability of picking exactly two red ones. The possible draws with exactly two red marbles are rrb, rbr, brr, rrg, rgr, grr. Then This is approximately 0.081. Pr(exactly two red marbles in three draws) ( ) 2 ( ) 2 2 6 2 =3 11 11 +3 3 11 11 = 108 1331. Section 6 Tree Diagrams 1. The tree diagram in Figure 1 shows the outcomes. We get the probabilities by multiplying numbers on edges of paths. the outcomes are payoffs of 0, 5, 10, 50 and 1200 dollars. From the edges of the tree, read: Pr(0) = 1 ( ) 1 = 1 4 2 8, Pr(5) = 1 ( ) 1 = 1 4 2 8, ( )( ) 1 1 Pr(10) = 4 = 1 4 2 2, Pr(50) = 1 ( ) 1 = 1 4 2 8, Pr(1200) = 1 ( ) 1 = 1 4 2 8. As a check, notice that these probabilities sum to 1. 13
50 5 10 10 10 10 10 1200 Figure 1: Tree diagram for Problem 1, Section 6. 14
3. From the tree diagram in Figure 2, obtain ( )( ) 1 1 Pr(Ford) = 2 = 1 6 3 9, Pr(Chevrolet) = 1 ( ) 1 = 1 6 3 18, Pr(VW) = 1 ( ) 1 = 1 6 2 12, ( )( ) 1 1 Pr(Porsche) = 2 = 1 6 2 6, Pr(Lamborghini) = 1 ( ) 1 = 1 6 2 12, Pr(tricycle) = 1 ( ) 1 = 1 6 2 12, ( )( ) 1 1 Pr(Mercedes) = 2 = 1 6 3 9, Pr(Honda) = 1 ( ) 1 = 1 6 3 18, Pr(tank) = 1 ( ) 1 = 1 6 2 12, ( )( ) 1 1 Pr(bicycle) = 3 = 1 6 4 8, Pr(Stanley Steamer) = 1 ( ) 1 = 1 6 4 24. 15
Ford 1/3 Ford 1/3 Chevrolet 1/3 VW Porsche 1/6 1/6 Lamborghini 1/6 tricycle 1/6 bicycle 1/6 1/6 Mercedes 1/3 Mercedes 1/3 1/3 Civic tank bicycle Porsche bicycle Stanley Steamer Figure 2: Tree diagram for Problem 3, Section 6. 16
years experience per cent of time products defective per cent as a decimal 10 1 0.01 5 to 10 years exp. 3 0.03 1 to 5 years exp. 5.7 0.057 < 1 year experience 11.3 0.113 Table 1: Information for Problem 1, Section 7. 5. From the tree diagram in Figure 3, we read the following probabilities: ( )( ) 1 1 Pr(0) = 5 = 1 5 4 4, ( )( ) 1 1 Pr(20) = 2 = 1 5 4 10, Pr(nickel) = 1 ( ) 1 = 1 5 4 20, Pr(chair) = 1 ( ) 1 = 1 5 4 20, ( )( ) 1 1 Pr(500) = 4 = 1 5 4 5, Pr(1000) = 1 ( ) 1 = 1 5 4 20, ( )( ) 1 1 Pr(1500) = 4 = 1 5 4 5, ( )( ) 1 1 Pr(lion) = 2 = 1 5 4 10. Section 7 Bayes s Theorem 1. The company hired people in four groupings. The table displays the given information. The table below displays these groupings by work experience, and, for each, the percentage this group constitutes of the total work force (written as a decimal), and a reiteration of the percentage (as a decimal) of products that are defective. By writing these percentages as decimals, they can be thought of as probabilities. For example, if 45 percent of the workforce is in one category, then the probability that a worker is in that category is taken to be 0.45. 17
0 0 house 1 0 chair 0 1/5 1/5 1/5 1/5 1/5 1000 0 house 2.05 500 500 500 house 3 500 1500 house 5 1500 1500 20 house 4 lion 1500 20 lion Figure 3: Tree diagram for Problem 5, Section 6. group per cent of work force (E j ) per cent of time products defective 10 0.45 0.01 5 to 10 years exp. 0.37 0.03 1 to 5 years exp. 0.07 0.057 < 1 year experience 0.11 0.113 Table 2: Percentages of workforce and products defective, by group (Problem 1, Section 7). 18
For each category or grouping, we want the probability that a worker is in this grouping, knowing that an item is defective. Let U be the event that an item is defective, and E j the event that a worker is in grouping j (column two of Table 2). We will use the equation Pr(E j U) Pr(E j )Pr(U E j ) = Pr(E 1 )Pr(U E 1 ) + Pr(E 2 )Pr(U E 2 ) + Pr(E 3 )Pr(U E 3 ) + Pr(E 4 )Pr(U E 4 ) for j =1, 2, 3, 4. The numbers Pr(U E j ) are read from column three of the last table. Now compute Pr(E 1 U) = (0.45)(0.01) (0.45)(0.01) +(0.37)(0.03) +(0.07)(0.057) + (0.11)(0.113) =0.141. This is the probability that a worker was in the ten or more years experience group, knowing that a product was defective. For j =2, 3, 4, the denominator in the Bayes formula is the same and we need only adjust the numerator (which is one term of the denominator). Compute and Pr(E 2 U) = (0.37)(0.03) (0.45)(0.01) +(0.37)(0.03) +(0.07)(0.057) + (0.11)(0.113) =0.347, Pr(E 3 U) = (0.07)(0.057 (0.45)(0.01) +(0.37)(0.03) +(0.07)(0.057) + (0.11)(0.113) =0.125 Pr(E 4 U) = (0.11)(0.0113) (0.45)(0.01) +(0.37)(0.03) +(0.07)(0.057) + (0.11)(0.113) =0.039. 3. Display the information of the problem: (a) Suppose a patient is selected. We want the probability that this patient is an adult man, if we are told that he or she survived at least two more years. Thus we want to compute Pr(adult man survived 2 years). 19
group survive < 1 year survive 2 years total group pop. adult men 619 257 876 adult women 471 320 791 boys 155 104 155 girls 190 152 190 Table 3: Information for Problem 3, Section 7. The total number of patients is 876 + 791 + 155 + 190 = 2012. The probability of choosing an adult man is therefore (to the best of our information) approximately 876/2012. Arguing similarly for the other probabilities in the general Bayes formula, we first obtain Pr(adult man was chosen survived 2 years) (190/2012)(35/190) (876/2012)(257/876) + (79012)(320/791) + (155/2012)(104/155) + (190/2012)(152/190) =0.309. (b) Now compute Pr(girl was chosen survived < 1 year) (190/2012)(38/190) (190/2012)(38/190) + (155/2012)(51/155) + (79012)(471/791) + (876/2012)(619/876) =0.032. 5. Let E j be the event that the gun is from city j in the table. Let U be the event that the gun explodes. The denominator in the general Bayes formula for this problem is (0.15)(0.02) + (0.07)(0.01) + (0.21)(0.06) +(0.05)(0.02) +(0.09)(0.03) +(0.44)(0.09) = 0.0594. Then Pr(E 1 U) = (0.15)(0.02) 0.0594 Pr(E 2 U) = (0.07)(0.01) 0.0594 Pr(E 3 U) = (0.21)(0.06) 0.0594 Pr(E 4 U) = (0.04)(0.02) 0.0594 Pr(E 5 U) = (0.09)(0.03) 0.0594 Pr(E 6 U) = (0.44)(0.09) 0.0594 =0.0505, =0.0118, =0.2125, =0.0135, =0.0455, =0.666. 20
Section 8 Expected Value 1. First compute the probability that at least three heads come up in seven flips. It is easier to work with the complementary event that fewer than three heads come up, which occurs if no heads, exactly one head, or exactly two heads come up. The number of ways this can happen is 7C 2 + 7 C 1 + 7 C 0 =29. Since there are 2 7 = 128 outcomes, then Pr(fewer than three heads) = 29 128 and Pr(at least three heads) = 1 29 128 = 99 128. The player wins five dollars if there are at least three heads. Therefore the player s expected value is 99 29 (5) 128 128 (9), which works out to about one dollar, eighty three cents. On average the play expects to win this amount each time the game is played. 3. Deal six cards, without regard to order. There are 52 C 6 =20, 358, 520 hands (assuming a 52 card deck). The number of hands with no aces is 48C 6 =12, 271, 512. Therefore which is approximately 0.603. Then Pr(no ace) = 12271512 20358520. Pr(at least one ace) = 1 12271512 20358520, which is approximately 0.397. For the player, the expected value is approximately (0.397)(45) (0.603)(30), which is approximately 0.225. The player can expect on average to lose between 22 and 23 cents per game. 5. There are 20 C 3 = 1140 ways to draw three marbles from the jar, disregarding the order of the draw. We need the probability that at least two are even-numbered. This happens if exactly two or all three have even numbers. For exactly two even, we can pick two of the ten even numbers in 10 C 2 ways, and with each choice, any of the ten odd numbers, 21
for 10( 10 C 2 ) ways of picking exactly two even-numbered marbles. We can pick exactly three even in 10 C 3 ways. Therefore the number of ways of picking two or three even numbered marbles is Therefore (10)( 10 C 2 )+ 10 C 3 = 570. Pr(two or three even numbered marbles come up) = 570 1140 = 1 2. The expected value to the player is ( ) ( ) 1 1 3 7 = 2. 2 2 On average, the player can expect to lose two dollars per game. 7. We need the probability that, in choosing four of the hats (without regard to order), we draw at least three red hats (meaning exactly two or exactly three red hats). If we draw all three red hats, then we must draw one blue hat, and there are four ways to do this. If we draw exactly two red hats, there are three ways to do this with the three red hats available, and then there are 4 C 2 ways of picking the other two hats from the four blue hats. This means that there are 3(6) = 18 ways of picking exactly two red hats. Therefore there are 22 ways of drawing at least two red hats, and Pr( 2 red hats) = 22 35. The payoff for the player is ( ) ( 22 (10) 1 22 ) (5), 35 35 which is about 4.42. The player can expect on average to win four dollars and forty two cents per game. 22