İ.D. Bilkent University Fall 2021-2022
Math255 Probability and Statistics
Midterm 1 Solutions
Problem 1. [6 pts] There are three coins each with possible outcomes heads (H) and
tails (T). Coin 1 is a fair coin with equally likely outcomes. Coins 2 and 3 are bent coins,
having probability of heads 5/6 and 1/6, respectively. Two of the three coins are picked
at random and each flipped once. Let A be the event that coin 1 is one of the two coins
picked. Let B be the event that the outcomes on the flipped coins are the same (both
heads or both tails).
(a) (3 pts) Compute P (B).
(b) (3 pts) Compute P (A|B).
Your solution should show your sample space Ω and identify the events A and B as subsets
of Ω.
(a) First, let us set up the probability model. We select the sample space as
Ω = {(1, 2, H, H), (1, 2, H, T ), (1, 2, T, H), (1, 2, T, T ), (1, 3, H, H), (1, 3, H, T ), (1, 3, T, H),
(1, 3, T, T ), (2, 3, H, H), (2, 3, H, T ), (2, 3, T, H), (2, 3, T, T )}
where an outcome (i, j, k, `) ∈ Ω corresponds to choosing coins i and j, obtaining k ∈
{H, T } on coin i, and ` ∈ {H, T } on coin j. The events A and B as subsets of Ω are given
by
A = {(1, 2, H, H), (1, 2, H, T ), (1, 2, T, H), (1, 2, T, T ), (1, 3, H, H), (1, 3, H, T ),
(1, 3, T, H), (1, 3, T, T )}
B = {(1, 2, H, H), (1, 2, T, T ), (1, 3, H, H), (1, 3, T, T ), (2, 3, H, H), (2, 3, T, T )}.
Let C12 = {(1, 2, H, H), (1, 2, H, T ), (1, 2, T, H), (1, 2, T, T )} denote the event that the cho-
sen coins are coins 1 and 2. Define C13 and C23 similarly. Since each pair of coins is
equally likely to be selected, we have P (C12 = P13 = P (C23 ) = 1/3. Furthermore, the sets
C12 , C13 , C23 form a partition of Ω. So, by the law of total probability, we have
P (B) = P (B|C12 )P (C12 ) + P (B|C13 )P (C13 ) + P (B|C23 )P (C23 )
1
= P (B|C12 ) + P (B|C13 ) + P (B|C23 ) .
3
We have
1 5 1 1 1
P (B|C12 ) = P {(1, 2, H, H), (1, 2, T, T )} C12 = · + · =
2 6 2 6 2
1 1 1 5 1
P (B|C13 ) = P {(1, 3, H, H), (1, 3, T, T )} C13 = · + · =
3 6 2 6 2
5 1 1 5 10 5
P (B|C23 ) = P {(2, 3, H, H), (2, 3, T, T )} C23 = · + · = =
6 6 6 6 36 18
Putting these together,
1 1 1 5 23
P (B) = + + = .
3 2 2 18 54
This study source was downloaded by 100000899606396 from CourseHero.com on 09-25-2025 13:47:22 GMT -05:00
https://www.coursehero.com/file/133311159/Midterm-1-2021-22-Fall-Solutionspdf/
Math255 Probability and Statistics
Midterm 1 Solutions
Problem 1. [6 pts] There are three coins each with possible outcomes heads (H) and
tails (T). Coin 1 is a fair coin with equally likely outcomes. Coins 2 and 3 are bent coins,
having probability of heads 5/6 and 1/6, respectively. Two of the three coins are picked
at random and each flipped once. Let A be the event that coin 1 is one of the two coins
picked. Let B be the event that the outcomes on the flipped coins are the same (both
heads or both tails).
(a) (3 pts) Compute P (B).
(b) (3 pts) Compute P (A|B).
Your solution should show your sample space Ω and identify the events A and B as subsets
of Ω.
(a) First, let us set up the probability model. We select the sample space as
Ω = {(1, 2, H, H), (1, 2, H, T ), (1, 2, T, H), (1, 2, T, T ), (1, 3, H, H), (1, 3, H, T ), (1, 3, T, H),
(1, 3, T, T ), (2, 3, H, H), (2, 3, H, T ), (2, 3, T, H), (2, 3, T, T )}
where an outcome (i, j, k, `) ∈ Ω corresponds to choosing coins i and j, obtaining k ∈
{H, T } on coin i, and ` ∈ {H, T } on coin j. The events A and B as subsets of Ω are given
by
A = {(1, 2, H, H), (1, 2, H, T ), (1, 2, T, H), (1, 2, T, T ), (1, 3, H, H), (1, 3, H, T ),
(1, 3, T, H), (1, 3, T, T )}
B = {(1, 2, H, H), (1, 2, T, T ), (1, 3, H, H), (1, 3, T, T ), (2, 3, H, H), (2, 3, T, T )}.
Let C12 = {(1, 2, H, H), (1, 2, H, T ), (1, 2, T, H), (1, 2, T, T )} denote the event that the cho-
sen coins are coins 1 and 2. Define C13 and C23 similarly. Since each pair of coins is
equally likely to be selected, we have P (C12 = P13 = P (C23 ) = 1/3. Furthermore, the sets
C12 , C13 , C23 form a partition of Ω. So, by the law of total probability, we have
P (B) = P (B|C12 )P (C12 ) + P (B|C13 )P (C13 ) + P (B|C23 )P (C23 )
1
= P (B|C12 ) + P (B|C13 ) + P (B|C23 ) .
3
We have
1 5 1 1 1
P (B|C12 ) = P {(1, 2, H, H), (1, 2, T, T )} C12 = · + · =
2 6 2 6 2
1 1 1 5 1
P (B|C13 ) = P {(1, 3, H, H), (1, 3, T, T )} C13 = · + · =
3 6 2 6 2
5 1 1 5 10 5
P (B|C23 ) = P {(2, 3, H, H), (2, 3, T, T )} C23 = · + · = =
6 6 6 6 36 18
Putting these together,
1 1 1 5 23
P (B) = + + = .
3 2 2 18 54
This study source was downloaded by 100000899606396 from CourseHero.com on 09-25-2025 13:47:22 GMT -05:00
https://www.coursehero.com/file/133311159/Midterm-1-2021-22-Fall-Solutionspdf/
