# Homework 1 Solution

Topics: Initialisms, Event, Probability Pages: 9 (1223 words) Published: October 31, 2012
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 that he is not a fool. Solution We need to ﬁnd P(B|Ac ). We know that

P(B|Ac ) =
=

0.7 − 0.55
=
1 − 0.6

P(B) − P(B ∩ A)
P(B ∩ Ac )
=
P(Ac )
1 − P(A)

0.375

2

Due on January 12, 2007

(1)

Solutions

First Problem Assignment

PROBLEM 3 (35 points) Express each of the following events in terms of the events A, B and C as well as the operations of complementation, union and intersection: (a) at least one of the events A, B , C occurs;

Solution

A∪B∪C
A

B
C

(b) at most one of the events A, B , C occurs;
Solution

[(A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A)]c
A

B
C

(c) none of the events A, B , C occurs;
Solution

Ac ∩ Bc ∩ Cc
A

B
C

3

Due on January 12, 2007

Solutions

First Problem Assignment

(d) all three events A, B , C occur;
Solution

A∩B∩C
A

B
C

(e) exactly one of the events A, B , C occurs;
Solution

[A ∩ (Bc ∩ Cc )] ∪ [B ∩ (Cc ∩ Ac )] ∪ [C ∩ (Ac ∩ Bc )] A

B
C

(f) events A and B occur, but not C ;
Solution

A ∩ B ∩ Cc
A

B
C

4

Due on January 12, 2007

Solutions

First Problem Assignment

(g) either event A occurs or, if not, then B also does not occur. Solution

A ∪ (Ac ∩ Bc )
A

B
C

In each case draw the corresponding Venn diagram.
PROBLEM 4 (14 points) Suppose A and B are two events with known probabilities. (a) Can you compute P(A ∪ B) in terms of P(A) and P(B)? If so, explain. If not, ﬁnd the largest and smallest possible values P(A ∪ B) can take in terms of P(A) and P(B) and give examples in which these values can be attained (i.e., give upper and lower bounds for P(A ∪ B).).

Solution No, P(A ∪ B) can not be computed in terms of P(A) and P(B) only. We know that,

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Knowing only P(A) and P(B) does not give complete information about P(A ∩ B). However we know that

P(A ∩ B)

0

and

P(A ∩ B)

max P(A), P(B)

min P(A), P(B)

P(A ∪ B)

P(A) + P(B)

The following cases can achieve the lower and the upper bound...

Please join StudyMode to read the full document