Homework 1 Solution

First Problem Assignment
EECS 401
Due on January 12, 2007
PROBLEM 1 (15 points) Fully explain your answers to the following questions.
(a) If events A and B are mutually exclusive and collectively exhaustive, are Ac and Bc mutually exclusive?
Solution Ac ∩ Bc = (A ∪ B)c = Ωc = ∅. Thus the events Ac and Bc are mutually exclusive. (b) If events A and B are mutually exclusive but not collectively exhaustive, are Ac and
Bc collectively exhaustive?
Solution Let C = (Ac ∪ Bc )c , that is the part that is not contained in Ac ∪ Bc . Using
De Morgan’s Law C = A ∩ B = ∅. Thus, there is nothing that is not a part of Ac or Bc .
Hence, Ac and Bc are collectively exhaustive.
(c) If events A and B are collectively exhaustive but not mutually exclusive, are Ac and
Bc collectively exhaustive?
Solution As in previous part, let C = (Ac ∪ Bc )c = A ∩ B which is not null. Thus, Ac and Bc are not collectively exhaustive.

PROBLEM 2 (10 points) Joe is a fool with probability 0.6, a thief with probability 0.7, and neither with probability 0.25.
(a) Determine the probability that he is a fool or a thief but not both.
Solution Let A be the event that Joe is a fool and B be the event that Joe is a thief. We are given that

P(A) = 0.6
P(B) = 0.7
P((A ∪ B)c ) = 0.25
This implies:

1

Due on January 12, 2007

Solutions

First Problem Assignment

P A ∪ B = 1 − P (A ∪ B)c = 0.75
P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.55
The event that he is a fool or a thief but not both is given by (A ∩ Bc ) ∪ (Ac ∩ B). Looking at the Venn diagram, the probability should be:

P (A ∩ Bc ) ∪ (Ac ∩ B) = P(A) + P(B) − 2P(A ∩ B) = 0.2

A

B

We can also derive this as follows

(A ∩ Bc ) ∪ (Ac ∩ B) = (A ∪ B) ∩ (A ∩ B)c
Thus,

P (A ∪ B) ∩ (A ∩ B)c = P(A ∪ B) + P (A ∩ B)c ) − P (A ∪ B) ∪ (A ∩ B)c
= P(A ∪ B) + 1 − P(A ∩ B) − P(S) = P(A ∪ B) − P(A ∩ B) = 0.2
This is the same expression as in (1).
(b) Determine the conditional probability that he is a thief, given