Solutions: Problems for Chapter 3 - Bryn Mawr College A: You are dealt ﬁve cards from a standard deck. Are you more likely to be dealt two pairs or three of a kind? Solutions: Problems

Solutions: Problems for Chapter 3


Problem A:

You are dealt five cards from a standard deck. Are you morelikely to be dealt two pairs or three of a kind?


experiment: choose 5 cards at random from a standard deck

Ω = 5-combinations of 52 cards

m(ω) = 1(52

5 )for all ω ∈ Ω


E = outcomes in Ω with three of a kind

P(E) =∑ω∈E

m(ω) =(number of outcomes in E)× 1(52

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.

n1 = 13, n2 =(4

3

), n3 =

(122

), n4 = 4, n5 = 4


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 tocomplete the task.

P(E) = N × 1(52

12)= 54,912

2,598,960 ≈ 0.0211↔ 2.11% chance


F = outcomes in Ω with two pairs

P(F ) =∑ω∈F

m(ω) =(number of outcomes in F )× 1(52

5 )


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 fromthose selected for the pairs.

n1 =(13

2

), n2 =

(42

), n3 =

(42

), n4 = 44


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 tocomplete the task.

P(F ) = N × 1(52

12)= 123,552

2,598,960 ≈ 0.0475↔ 4.75% chance


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


Problem B:

You decide to play Roulette. Are you more likely to win at leastonce in 10 bets on a column or in 5 bets on red?


Sub-Problems:

(1) one bet on a column

(2) ten bets on a column

(3) one bet on red

(4) five bets on red


(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) = 138 for all ω ∈ Ω

P(E) =∑ω∈E

m(ω) = (number of outcomes in E) × 138 = 12

38


(2) ten bets on a column

Bernoulli trials process

n = 10

experiment: one bet on a column

success: winfailure: lose

p = 1238 , q = 26

38

X = number of successes


P(X ≥ 1) = 1− P(X = 0)

= 1− b(10, 12

38 ,0)

= 1−(10

0

) (1238

)0 (2638

)10

− 1−(26

38

)10

≈ 0.976↔ 97.6%


(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) = 138 for all ω ∈ Ω

P(F ) =∑ω∈F

m(ω) = (number of outcomes in F ) × 138 = 18

38


(4) five bets on red

Bernoulli trials process

n = 5

experiment: one bet on red

success: winfailure: lose

p = 1838 , q = 20

38

X = number of successes


P(X ≥ 1) = 1− P(X = 0)

= 1− b(10, 18

38 ,0)

= 1−(10

0

) (1838

)0 (2038

)10

− 1−(20

38

)10

≈ 0.957↔ 95.7%


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


Problem C:

You conduct the following experiment in our class of 25students:

Everyone privately flips their own coin. If it lands heads, theyanswer “yes” to the question you give them. If it lands tails, theyanswer the question truthfully.

The question is, “Have you ever cheated on an exam incollege?”

How many “yes” answers out of 25 would convince you thatthere are cheaters in the class?


Set this experiment up as a hypothesis test using Example 3.11as 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 = 12

alternative hypothesis: p > 12


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 assumptionthat the null hypothesis is true. In other words, we give studentsthe benefit of the doubt, and assume there are no cheaters inthe class. We reject this assumption only if the evidence isstrong enough.


(P ≥ m) =25∑

k=m

b(25, 1

2 , k)

=25∑

k=m

(25k

)(12

)k (12

)(25−k)

=25∑

k=m

(25k

)(12

)25


The Mathematica output below suggests we choose m = 18.

TableB:m, NBâk=m

25Binomial@25, kD 1

2

25F>, 8m, 13, 25<F

9813, 0.5<, 814, 0.345019<, 815, 0.212178<, 816, 0.114761<,817, 0.0538761<, 818, 0.0216426<, 819, 0.00731665<,820, 0.00203866<, 821, 0.00045526<, 822, 0.0000782609<,923, 9.71556 ´ 10-6=, 924, 7.7486 ´ 10-7=, 925, 2.98023 ´ 10-8==

18 or more “yes” answers would convince us there are cheatersin the class.



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Solutions: Problems for Chapter 3 - Bryn Mawr College A: You are dealt ﬁve cards from a standard deck. Are you more likely to be dealt two pairs or three of a kind? Solutions: Problems