Solutions: Problems for Chapter 3. Solutions: Problems for Chapter 3

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1

2 Problem A: You are dealt five cards from a standard deck. Are you more likely to be dealt two pairs or three of a kind?

3 experiment: choose 5 cards at random from a standard deck Ω = {5-combinations of 52 cards} m(ω) = ( 1 for all ω Ω 52 5 )

4 E = {outcomes in Ω with three of a kind} P(E) = ω E m(ω) =(number of outcomes in E) 1 ( 52 5 )

5 Task: Create a hand with three of a kind. Stage 1: Choose the denomination (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, or K) for the three of a kind. Stage 2: Choose 3 cards from the selected denomination. Stage 3: Choose the denominations for the remaining 2 cards. Stage 4. Choose one card from the smaller denomination. Stage 5. Choose one card from the larger denomination. n 1 = 13, n 2 = ( ) 4 3, n3 = ( ) 12 2, n4 = 4, n 5 = 4

6 The number of ways to complete the task is: N = 13 ( 4 3) ( 12 2 ) 4 4 = 54, 912. The number of outcomes in E is the number N of ways to complete the task. P(E) = N ( ) = 54,912 2,598, % chance

7 F = {outcomes in Ω with two pairs} P(F) = ω F m(ω) =(number of outcomes in F) 1 ( 52 5 )

8 Task: Create a hand with two pairs. Stage 1: Choose two denominations (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, or K) for the pairs. Stage 2: Choose 2 cards from the smaller denomination. Stage 3: Choose 2 cards from the larger denomination. Stage 4. Choose one card with a denomination different from those selected for the pairs. n 1 = ( ) 13 2, n2 = ( ) 4 2, n3 = ( 4 2), n4 = 44

9 The number of ways to complete the task is: N = ( 13 2 ) ( 4 2) ( 4 2) 44 = 123, 552. The number of outcomes in F is the number N of ways to complete the task. P(F) = N ( ) = 123,552 2,598, % chance

10 Your are more likely to be dealt two pairs (4.75% chance) than three of a kind (2.11% chance).

11 Problem B: You decide to play Roulette. Are you more likely to win at least once in 10 bets on a column or in 5 bets on red?

12 Sub-Problems: (1) one bet on a column (2) ten bets on a column (3) one bet on red (4) five bets on red

13 (1) one bet on a column experiment: spin a Roulette wheel one time Ω = {0, 00, 1, 2, 3,..., 36} E: win a bet on a column m(ω) = 1 38 for all ω Ω P(E) = ω E m(ω) = (number of outcomes in E) 1 38 = 12 38

14 (2) ten bets on a column Bernoulli trials process n = 10 experiment: one bet on a column success: win failure: lose p = 12 38, q = X = number of successes

15 P(X 1) = 1 P(X = 0) = 1 b ( 10, 12 38, 0) = 1 ( 10 0 ) ( ( ) % ) 0 ( 26 ) 10 38

16 (3) one bet on red experiment: spin a Roulette wheel one time Ω = {0, 00, 1, 2, 3,..., 36} F : win a bet on red m(ω) = 1 38 for all ω Ω P(F) = m(ω) = (number of outcomes in F) 1 38 = ω F

17 (4) five bets on red Bernoulli trials process n = 5 experiment: one bet on red success: win failure: lose p = 18 38, q = X = number of successes

18 P(X 1) = 1 P(X = 0) = 1 b ( 10, 18 38, 0) = 1 ( 10 0 ) ( ( ) % ) 0 ( 20 ) 10 38

19 You are slightly more likely to win at least once in 10 bets on a column (97.6% chance) than you are to win at least once in 5 bets on red (95.7% chance).

20 Problem C: You conduct the following experiment in our class of 25 students: Everyone privately flips their own coin. If it lands heads, they answer yes to the question you give them. If it lands tails, they answer the question truthfully. The question is, Have you ever cheated on an exam in college? How many yes answers out of 25 would convince you that there are cheaters in the class?

21 Set this experiment up as a hypothesis test using Example 3.11 as a model. null hypothesis: there are no cheaters in the class alternative hypothesis: there are some cheaters in the class p = probability of a yes response null hypothesis: p = 1 2 alternative hypothesis: p > 1 2

22 n = 25 people answer the question X = number of yes responses m = critical value We reject the null hypothesis if X m, and accept it if X < m. We choose m so that P(X m) < 0.05 under the assumption that the null hypothesis is true. In other words, we give students the benefit of the doubt, and assume there are no cheaters in the class. We reject this assumption only if the evidence is strong enough.

23 (P m) = = = 25 k=m 25 k=m 25 k=m b ( 25, 1 2, k) ( 25 k ( 25 k ) ( 1 2 ) ( 1 2 ) k ( ) 1 (25 k) 2 ) 25

24 The Mathematica output below suggests we choose m = Table m, N Binomial 25, k 1 2 k m 25, m, 13, 25 13, 0.5, 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, or more yes answers would convince us there are cheaters in the class.

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