Problem B.2
Question: Let A, B, and C be subsets of the set Ω. Describe the following
sets using unions, intersections, and complements:
(a) The set of elements that are in each of the three sets.
(b) The set of elements that are in A but neither in B nor in C.
(c) The set of elements that are in at least one of the sets A or B.
(d) The set of elements that are in both A and B but not in C.
(e) The set of elements that are in A, but not in B or C, or both.
Work:
(a) The set of elements that are in each of the three sets is given by:
A∩B∩C
(b) The set of elements that are in A but neither in B nor in C is:
A ∩ Bc ∩ C c
(c) The set of elements that are in at least one of the sets A or B is:
A∪B
(d) The set of elements that are in both A and B but not in C is:
A ∩ B ∩ Cc
(e) The set of elements that are in A, but not in B or C, or both, is:
A ∩ (B ∪ C)c
Answer:
(a) A ∩ B ∩ C
(b) A ∩ B c ∩ C c
(c) A ∪ B
(d) A ∩ B ∩ C c
(e) A ∩ (B ∪ C)c
1
Problem C.10
Question: Find the number of poker hands that fall into the following categories:
(a) Two pairs (Two cards of the same rank, two cards of another rank, and a
fifth card with a different rank).
(b) Three of a kind (Three cards of the same rank with two cards that have
ranks different from each other and the first three).
(c) Straight (Five cards with ranks in sequence, not in the same suit).
(d) Flush (Five cards of the same suit, not in sequential rank).
(e) Full house (Three cards of one rank and two cards of another rank).
(f) Four of a kind (Four cards of the same rank and another card).
(g) Straight flush (Five cards in the same suit and in sequence).
Work:
(a) Two pairs:
13
Ways to choose two ranks for the pairs =
2
4
4
Ways to choose two suits for each pair =
×
2
2
11
4
Ways to choose the rank and suit for the fifth card =
×
1
1
13
4
4
11
4
Total =
×
×
×
×
= 123, 552
2
2
2
1
1
(b) Three of a kind:
13
1
4
Ways to choose 3 suits for the three cards =
3
12
4
4
Ways to choose 2 other different ranks and suits =
×
×
2
1
1
13
4
12
4
4
Total =
×
×
×
×
= 54, 912
1
3
2
1
1
Ways to choose the rank of the three cards =
2
(c) Straight:
Ways to choose a straight sequence = 10 × 45 − 40 = 10, 200
(d) Flush:
13
Ways to choose a flush = 4 ×
− 40 = 5, 108
5
(e) Full house:
13
1
4
Ways to choose the suits for the three cards =
3
12
Ways to choose the rank of the pair =
1
4
Ways to choose 2 suits for the pair =
2
13
4
12
4
Total =
×
×
×
= 3, 744
1
3
1
2
Ways to choose the rank of the three cards =
(f) Four of a kind:
13
Ways to choose the rank of the 4 cards =
1
12
4
Ways to choose the 5th card =
×
1
1
13
4
12
4
Total =
×
×
×
= 624
1
4
1
1
(g) Straight flush:
Ways to choose a straight flush = 10 × 4 = 40
Problem 1.4
Question: A kindergarten class randomly chooses one of the 50 state flags each
day on Monday, Tuesday, and Wednesday. What are the following probabilities?
(a) Describe a sample space S and a probability measure P for the experiment.
(b) What is the probability that Wisconsin’s flag is hung on Monday, Michigan’s flag on Tuesday, and California’s flag on Wednesday?
3
(c) What is the probability that Wisconsin’s flag will be hung at least two of
the three days?
Work:
(a) The sample space consists of all possible sequences of three flags chosen
from 50, and the total number of outcomes is 503 = 125, 000.
(b) The probability of Wisconsin on Monday, Michigan on Tuesday, and California on Wednesday is:
P (specific sequence) =
1
125, 000
(c) The probability of Wisconsin being chosen at least two days is:
P (Wisconsin on at least 2 days) = 1 − P (Wisconsin on 0 or 1 day)
Compute:
49
50
3
1
×
50
P (Wisconsin on 0 days) =
P (Wisconsin on 1 day) = 3 ×
49
50
2
P (Wisconsin on at least 2 days) = 1 − [P (0 days) + P (1 day)]
Answer:
(a) S = 503 , P = 5013
1
(b) 125,000
148
(c) P (Wisconsin on at least 2 days) = 125,000
= 0.001184
Problem 1.6
Question: We have an urn with 3 green and 4 yellow balls. We choose 2 balls
randomly without replacement. Let A be the event that we have two differentcolored balls in our sample.
(a) Describe a possible sample space and the event A.
(b) Compute P (A).
Work:
(a) The total number of ways to choose 2 balls from 7 is 72 = 21. The event
A consists of choosing one green and one yellow ball, which can occur in
3 × 4 = 12 ways.
4
(b) The probability of event A is:
P (A) =
12
4
=
21
7
Answer:
(a) The sample space consists of
outcomes.
7
2
= 21 outcomes, and A has 12 favorable
(b) P (A) = 74
Problem 1.12
Question: We roll a fair die repeatedly until we see the number four appear
and then we stop.
(a) What is the probability that we need at most 3 rolls?
(b) What is the probability that we need an even number of die rolls?
Work:
(a) The probability of rolling a 4 on a fair die is P (rolling a 4) = 16 .
Probability of needing 1, 2, or 3 rolls:
1
P (1 roll) =
6
5
1
P (2 rolls) =
×
6
6
2
1
5
×
P (3 rolls) =
6
6
Adding these probabilities:
1
P (at most 3 rolls) = +
6
2
5
1
5
1
× +
×
6
6
6
6
1
5
25
+
+
6 36 216
6
5
25
=
+
+
36 36 216
11
25
=
+
36 216
66
25
91
=
+
=
216 216
216
=
5
(b) The probability of needing 2 rolls, 4 rolls, 6 rolls, and so on, forms a
geometric series where:
1
5
× ,
P (2 rolls) =
6
6
3
1
5
× ,
P (4 rolls) =
6
6
This is a geometric series with first term a =
2
ratio r = 56 .
5
6
5
1
5
× ,...
P (6 rolls) =
6
6
5
and common
× 16 = 36
The sum of this infinite geometric series is given by:
P (even number of rolls) =
5
a
36
=
2
1−r
1− 5
6
=
5
36
5
1 − 25
36
Answer:
(a) ≈ 0.4213
(b) ≈ 0.4545
6
= 36
11 =
36
5
11