EE351K: Probability and Random Processes Homework #2 Solutions
Dr. Pedro Santacruz Due: June 20, 2022
Homework #2 Solutions
You should solve these problems with your homework group. I recommend that you start early
and get help in office hours if needed. Only submit one copy to Gradescope and ensure all team
members are attached to the submission.
Problem 1: Dice
A fair six-sided die is thrown twice. Let B be the event that the first number thrown is not larger
than 3, and let C be the event that the sum of the two numbers thrown equals 5.
(i) Find the probabilities P(B) and P(C).
(ii) Find the conditional probabilities of P(C | B), and of P(B | C).
Solution
(i) The event B can be written as B = {(x, y) | x ∈ {1, 2, 3}, y ∈ {1, 2, 3, 4, 5, 6}}.
The event C can be described as C = {(x, y) | x + y = 5; x, y ∈ {1, 2, 3, 4, 5, 6}}.
Clearly, B has 18 elements and C has 4 elements.
∴, P(B) = 18 1
36 = 2 and P(C) = 364
= 19 .
(ii) The only common elements in B and C are (1, 4), (2, 3) and (3, 2).
P(B ∩ C)
∴, P(C | B) =
P(B)
(3)
1
= (36
1
) =
2
6
and
P(B ∩ C)
P(B | C) =
P(B)
(3)
3
= (36
1
) =
9
4
Problem 2: Base Rate Fallacy
A hypothetical country is looking to buy a new RADAR system to tighten it’s border security.
After a lot of market research, it has narrowed it’s options down to one of 2 RADARs with the
following specifications:
(i) RADAR A:
(a) Probability of issuing an alert given an enemy is present is 99%
(b) Probability of issuing an alert given no enemy is present is 10%
(ii) RADAR B:
Dr. Pedro Santacruz Due: June 20, 2022
Homework #2 Solutions
You should solve these problems with your homework group. I recommend that you start early
and get help in office hours if needed. Only submit one copy to Gradescope and ensure all team
members are attached to the submission.
Problem 1: Dice
A fair six-sided die is thrown twice. Let B be the event that the first number thrown is not larger
than 3, and let C be the event that the sum of the two numbers thrown equals 5.
(i) Find the probabilities P(B) and P(C).
(ii) Find the conditional probabilities of P(C | B), and of P(B | C).
Solution
(i) The event B can be written as B = {(x, y) | x ∈ {1, 2, 3}, y ∈ {1, 2, 3, 4, 5, 6}}.
The event C can be described as C = {(x, y) | x + y = 5; x, y ∈ {1, 2, 3, 4, 5, 6}}.
Clearly, B has 18 elements and C has 4 elements.
∴, P(B) = 18 1
36 = 2 and P(C) = 364
= 19 .
(ii) The only common elements in B and C are (1, 4), (2, 3) and (3, 2).
P(B ∩ C)
∴, P(C | B) =
P(B)
(3)
1
= (36
1
) =
2
6
and
P(B ∩ C)
P(B | C) =
P(B)
(3)
3
= (36
1
) =
9
4
Problem 2: Base Rate Fallacy
A hypothetical country is looking to buy a new RADAR system to tighten it’s border security.
After a lot of market research, it has narrowed it’s options down to one of 2 RADARs with the
following specifications:
(i) RADAR A:
(a) Probability of issuing an alert given an enemy is present is 99%
(b) Probability of issuing an alert given no enemy is present is 10%
(ii) RADAR B:
