Lenarz Math 102 Practice Exam # 3 Name: 1. A 10-sided die is rolled 100 times with the following results:

Transcription

1 Lenarz Math 102 Practice Exam # 3 Name: 1. A 10-sided die is rolled 100 times with the following results: Outcome Frequency (a) What is the experimental probability of rolling a 5? P (5) (b) What is the theoretical probability of rolling a 5? P (5) (c) What is the experimental probability of rolling a multiple of 3? P (multiple of 3) (d) What is the theoretical probability of rolling a multiple of 3? P (multiple of 3)

2 Math 102 Page 2 2. One card is drawn from a standard 52 card deck. What is the probability of drawing: (a) the queen of hearts? P (queen of hearts) 1 52 (b) a queen? P (queen) (c) a heart? P (heart) (d) a face card (J, Q, or K)? P (face card) Two -sided dice are rolled. What is the probability of rolling (a) a total of? There are 5 ways to get a total of {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} so we have P (total of ) 5 3 (b) not a total of? This is simply the complement of the event in part (a), so we have P (not a total of ) 1 P (total of )

3 Math 102 Page 3 (c) total of or 7? The keyword or means we are looking for the probability of a union, namely P (total of total of 7). To find this we need P (total of ), P (total of 7), and P (total of total of 7). We computed P (total of ) in part (a). There are ways to get a total of 7 {(1, ), (2, 5), (3, 4), (4, 3), (5, 2), (, 1)} so we have P (total of 7) 3 1 Also note that P (total of total of 7) 0 since it is impossible to have a total of and a total of 7 simultaneously. Thus we have P (total of total of 7) P (total of ) + P ( total of 7) P (total of total of 7) (d) total of or more? The probability of getting a total of or more is the complement of the probability of getting a total of 2, 3, 4, or 5. There is only one way to get a total of 2, 2 ways to get a total of 3, 3 ways to get a total of 4 and 4 ways to get a total of 5. Also, the four events (getting a total of 2, a total of 3, a total of 4 and total of 5) are mutually exclusive. Thus we have P (total of or more) 1 P (total of 2, 3, 4, or 5) ( 1 P (total of 2) + P (total of 3) ) +P (total of 4) + P (total of 5) ( )

4 Math 102 Page 4 4. A gumball machine has gumballs of five flavors. There are 10 apple, 15 berry, 12 cherry, 8 orange, and 9 mint. When a quarter is put into the machine, it dispenses 5 gumballs at random. What is the probability that (a) each gumball is a different flavor? There are gumballs in the machine and 3, 12, 510 ways to choose any 5 of the gumballs. There are 54C 5 54! 39!5! 10C 1 15 C 1 12 C 1 8 C 1 9 C , 00 ways to choose one of each flavor. So the probability of selecting one gumball of each flavor is P (one of each flavor) 10 C 1 15 C 1 12 C 1 8 C 1 9 C 1 129, 00 3, 12, , C 5 (b) at least two gumballs are the same flavor? This is the complement of the event in part (a), so we have P (at least 2 same flavor) 1 P (one of each) , , , A coin is tossed ten times in a row. Find the probability that (a) no tails show. The probability that no tails show is the same as the probability that all heads show. This is the probability that the first coin is heads AND the second coin is heads AND... so we have an intersection of independent events.

5 Math 102 Page 5 Thus P (no tails) P (10 heads) }{{} 10 times , 024 (b) exactly three tails show. The probability of getting heads is 1 and the probability of getting 2 tails is 1. There are 2 10C 3 10! 120 ways to have exactly three tails (you 7!3! need to choose which 3 of the 10 tosses will be tails, the rest will be heads). So the probability of getting exactly 3 tails is P (exactly 3 tails) ( ) ( ) C (c) toss exactly twice as many heads as tails. To have twice as many heads as tails we could have 1 tail and 2 heads (3 coins total), 2 tails and 4 heads ( coins total), 3 tails and heads (9 coins total), 4 tails and 8 heads (12 coins total), etc. Notice there is no way to have twice as many heads as tails and get 10 coins. So it is impossible to do! Thus P (twice as many heads as tails) 0. A group of 15 students is to be split into 3 groups of 5. In how many ways can this be done? There are 15 C 5 ways to choose the first group, 10 C 5 ways to choose the second group and 5 C 5 ways to choose the third group. So we multiply to get the number of ways to choose all three groups. But that overcounts because it takes the order of the groups into account (the first group is treated as different than the second group, etc). So we need to divide out by the number of ways we can order

6 Math 102 Page the 3 groups which is 3!. So we have 15C 5 10 C 5 5 C 5 3! 15! 10!5! 10! 5! 5!5! 5!1! 3! ! , 75 12, Five cards are drawn from a standard 52 card deck. What is the probability of drawing (a) 5 cards of the same color? P (5 of same color) , 998 (b) a full house (3 cards of one rank and two cards of a different rank)? The number of ways to draw any 5 cards is 52 C 5 2, 598, 90. The number of ways to get a full house is # ways to # of ways to # of ways to # of ways to pick rank for pick suits for pick rank pick suits 3 of a kind 3 of a kind for pair for pair 13C 1 4C 3 12C 1 4C 2 which is So the probability of drawing a full house is P (full house)

7 Math 102 Page 7 8. Suppose a jar has 4 coins: a penny, a nickel, a dime and a quarter. You remove two coins at random without replacement. Let A be the event you remove the quarter. Let B be the event you remove the dime. Let C be the event you remove less than 12 cents. (a) List the sample space. S {(p, n), (p, d), (p, q), (n, p), (n, d), (n, q), (d, p), (d, n), (d, q), (q, p), (q, n), (q, d)} (b) Draw a probability tree diagram to represent the possible scenarios. (c) Which pair(s) of theses events is (are) mutually exclusive? A {(p, q), (n, q), (d, q), (q, p), (q, n), (q, d)} B {(p, d), (n, d), (d, p), (d, n), (d, q), (q, d)} C {(p, n), (p, d), (n, p), (d, p)} So A B {(d, q), (q, d)} A C B C {(p, d), (d, p)} Thus only A and C are mutually exclusive.

8 Math 102 Page 8 (d) Find P (A), P (C), and P (B). P (A) P (C) P (B) (e) Compute and interpret P (A B) and P (B C). P (A B) is the probability that either A or B occurs, that is the probability that either a quarter or dime is removed. P (A B) P (A) + P (B) P (A B) P (B C) is the probability that either B or C occurs, that is the probability that either a dime or less than 12 cents is removed. P (B C) P (B) + P (C) P (B C) A dental assistant randomly sampled 200 patients and classified them according to whether or not they had a least one cavity in their last checkup and according to what type of tooth decay preventative measures they used. The information is as follows

9 Math 102 Page 9 At least one cavity No Cavities Brush only 9 2 Brush and floss only Brush and tooth sealants only Brush, floss and tooth sealants 3 4 If a patient is picked at random from this group, find the probability that (a) the patient had a least one cavity. The number of patients with at least one cavity is and the total number of patients is 200 so P (at least one cavity) (b) the patient brushes only. The number of patients that brush only is and the total number of patients is 200 so P (only brush) (c) the patient had no cavities, given s/he brushes, flosses and has tooth sealants. The number of patients that have no cavities and brush, floss and use sealants is 4. So the probability a patient has no cavities and brushes, flosses and uses sealants is P (no cavities brush, floss, and sealant) The number of patients that brush, floss and use sealants is so the probability a patient brushes, flosses and uses sealants is P (brush, floss, and sealants)

10 Math 102 Page 10 Thus the probability that a patient has no cavities, given s/he brushes, flosses and uses sealants is P (no cavities brush, floss, and sealants) P (no cavities brush, floss, and sealants) P (brush, floss, and sealants) (d) the patient brushes only, given that s/he had at least one cavity. The number of patients that only brush and have at least one cavity is 9. So the probability a patient only brushes and has at least one cavity is P (only brushes at least one cavity) The number of patients that have at least one cavity is so the probability a patient has at least one cavity is P (at least one cavity) Thus the probability that a patient brushes only, given s/he has at least one cavity is P (brushes only at least one cavity) P (brushes only at least one cavity) P (at least one cavity) A small college has two calculus classes. The first class has 25 students, 15 of whom are female, and the other class has 18 students, 8 of whom are female. One of the classes is selected at random and then two students are randomly selected from the class for an interview. If both of the students are female, what is the probability they came from the first class?

11 Math 102 Page 11 Call the first class Class A and the second class Class B. Then we have the following data So we get the following probability tree Female Male Class A Class B 8 10 From the tree we see that the probability of choosing class A and choosing two females is P (A 2 females) The probability of choosing two females is P (2 females) Thus the probability of choosing class A, given that two females were chosen is P (A 2 females) P (A 2 females) P (2 females) The probability is 0. that a student will study for an exam. If the student studies, she has a 0.8 chance of getting an A on the exam. If she does not study, she has a 0.3 probability of getting an A. Make a probability tree for this situation. What is the probability that she gets an A? If she gets an A, what is the conditional probability that

12 Math 102 Page 12 she studied? We get the following probability tree (where S denotes study, N denotes not study, A denotes gets an A and NA denotes does not get an A) From the tree we see that P (A) We also know that P (S A) so we have P (S A) P (S A) P (A)

13 Math 102 Page If you consider the value of a roll of a single -sided die to be the number that is rolled, what is the expected value of the roll of a single die? E ( ) 1 (1) + ( ) 1 (2) + ( ) 1 (3) + ( ) 1 (4) + ( ) 1 (5) + ( ) 1 () Consider a game that consists of drawing a single card at random from a standard 52 card deck. You pay $3 to play the game and the $3 is not returned. If you draw an ace you win $10. If you draw a king or a queen, you win $5. How much should you expect to win or lose on average if you play this game? E ( ) ( ) 4 8 (10 3) + (5 3) ( ) ( ) ( (7) + (2) ) ( 3) ( ) 40 (0 3) 52 You would expect to loose $1.4.

14 Math 102 Page Suppose you just inherited $100, 000 and you are trying to decide how to invest it for the next year. You have narrowed it down to two choices. The first choice is to invest it in a bond with a guaranteed return of 5% interest at the end of the year. The second choice is to bet it all on the Super Bowl with a 51% chance of doubling your money and a 49% chance of losing it all. Which investment option has the highest expected value? The bond has an expected value of E (1)(100, , ) 105, 000 Betting on the Super Bowl has an expected value of E (0.51)(200, 000) + (0.49)(0) 102, 000 So the bond has the higher expected value.