Math 255 Probability and Statistics Midterm 1 Solutions

Bilkent University
Spring 2020-21
Math 255 Probability and Statistics
Midterm 1, March 8, 2021
Solutions


1. [6 pts] Let A, B, C be three independent events in a probability space (Ω, P ) with P (A) = 0.2,
P (B) = 0.3, and P (C) = 0.4.
Compute the following probabilities. Each part is 2 pts.
(a) P ((A ∪ B) ∩ C c )
(b) P (B c ∪ C c | Ac )
(c) P (A ∩ B | B ∩ C)

Solution.
(a) The main point of this part is to observe that if A, B, C are independent events then A ∪ B and
C c are independent.

P ((A ∪ B) ∩ C c ) = P (A ∪ B)P (C c )
= [P (A) + P (B) − P (AB)] P (C c )
= [P (A) + P (B) − P (A)P (B)] P (C c ) (independence of A and B)
= (0.2 + 0.3 − (0.2)(0.3))(1 − 0.4) = (0.44)(0.6) = 0.264

It is also instructive to solve this problem from first principles.

P ((A ∪ B) ∩ C c ) = P (AC c ∪ BC c )
= P (AC c ) + P (BC c ) − P (AC c BC c )
= P (AC c ) + P (BC c ) − P (ABC c )
= [P (A) − P (AC)] + [P (B) − P (BC)] − [P (AB) − P (ABC)]
= [P (A) − P (A)P (C)] + [P (B) − P (B)P (C)]
− [P (A)P (B) − P (A)P (B)P (C)] (by independence of A, B, C)
= [P (A)(1 − P (C))] + [P (B)(1 − P (C))] − [P (A)P (B)(1 − P (C))]
= [P (A) + P (B) − P (A)P (B)] (1 − P (C)) = 0.264 (as above)

(b) We now use tha fact that B c ∪ C c and Ac are independent events.

P (B c ∪ C c | Ac ) = P (B c ∪ C c )
= P (B c ) + P (C c ) − P (B c C c )
= P (B c ) + P (C c ) − P (B c )P (C c ) (by independence of B c and C c )
= (1 − 0.3) + (1 − 0.4) − (1 − 0.3)(1 − 0.4)
= 0.7 + 0.6 − (0.7)(0.6) = 1.3 ∗ 0.42 = 0.880.

(c) Now we have to be careful since (A ∩ B) and (B ∩ C) are not independent events, as they have
B in common.
P ((A ∩ B) ∩ (B ∩ C))
P (A ∩ B | B ∩ C) =
P (B ∩ C)
P (A ∩ B ∩ C)
=
P (B ∩ C)
P (A)P (B)P (C)
= (independence)
P (B)P (C)
= P (A) = 0.2




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