Question

1. Suppose that A. B. and C are events such that PA-PlB) = 0.3. PC] = 0.55, PAT B) = For each of the events given below in parts (a)-(d), do the following: (i) Write a set expression for the event. (Note that there are multiple ways to write this in many cases.) (ii) Evaluate the probability of the event. Hint: Draw the Venn Diagram. You may then want to identify the probabilities of each of the disjoint regions in the diagram before starting the problem.) (a) At least one of the events A, B, or C occurs (b) Exactly one of A, B, or C occurs. (c) At most one of A, B, or C occurs. (d) C occurs, but neither A nor B occurs.

Answer 1

(a)

P(at least one of the events A,B, or C occurs) = 1 - P(none of the events A,B, and C occur) = 1 - P(A^c$\cap$B^c$\cap$C^c) = 1 - 0.1 = 0.9

.

(b)

P(exactly one of A,B, or C occurs) = P(A$\cap$B^c$\cap$C^c) + P(A^c$\cap$B$\cap$C^c) + P(A^c$\cap$B^c$\cap$C)

• P(A^c$\cap$B^c) = 1 - P(A$\cup$B) = 1 - [P(A)+P(B)-P(A$\cap$B)] = 1 - [0.3+0.3] = 0.4

P(A^c$\cap$B^c$\cap$C) = P(A^c$\cap$B^c) - P(A^c$\cap$B^c$\cap$C^c) = 0.4-0.1 = 0.3

• P(A^c$\cap$C^c) = P(C^c) - P(A$\cap$C^c) = 1 - P(C) - P(A$\cap$C^c) = 1-0.55-0.2 = 0.25

P(A^c$\cap$B$\cap$C^c) = P(A^c$\cap$C^c) - P(A^c$\cap$B^c$\cap$C^c) = 0.25-0.1 = 0.15

• P(A$\cup$B$\cup$C) = 1 - P(A^c$\cap$B^c$\cap$C^c) = 1-0.1 = 0.9

P(A$\cup$B$\cup$C) = P(A$\cup$B) + P(C) - P[(A$\cup$B)$\cap$C] = [P(A)+P(B)-P(A$\cap$B)] + P(C) - P[(A$\cup$B)$\cap$C] = 0.3+0.3+0.55 - P[(A$\cup$B)$\cap$C] $\Rightarrow$ P[(A$\cup$B)$\cap$C] = 0.3+0.3+0.55 - P(A$\cup$B$\cup$C) = 1.15-0.9 = 0.25

P[(A$\cup$B)$\cap$C] = P[(A$\cap$C)$\cup$(B$\cap$C)] = P(A$\cap$C) + P(B$\cap$C) - P(A$\cap$B$\cap$C) = P(A) - P(A$\cap$C^c) + P(B$\cap$C) - P(A$\cap$B$\cap$C)

$\Rightarrow$ 0.25 = 0.3-0.2 + P(B$\cap$C) - 0 { $\because$ P(A$\cap$B)=0$\Rightarrow$P(A$\cap$B$\cap$C)=0 }   

$\Rightarrow$ P(B$\cap$C) = 0.25-0.3+0.2 = 0.15

P(B^c$\cap$C^c) = 1 - P(B$\cup$C) = 1 - [P(B)+P(C)-P(B$\cap$C)] =1 - [0.3+0.55-0.15] = 0.3

P(A$\cap$B^c$\cap$C^c) = P(B^c$\cap$C^c) - P(A^c$\cap$B^c$\cap$C^c) = 0.3-0.1 = 0.2

$\therefore$ P(exactly one of A,B, or C occurs) = P(A$\cap$B^c$\cap$C^c) + P(A^c$\cap$B$\cap$C^c) + P(A^c$\cap$B^c$\cap$C) = 0.2+0.15+0.3 = 0.65

.

(c)

P(at most one of A,B, or C occurs) = P(none of A,B and C occur) + P(exactly one of A,B, or C occurs)

= P(A^c$\cap$B^c$\cap$C^c) + 0.65 = 0.1+0.65 = 0.75

.

(d)

P(C occurs, but neither A nor B occurs) = = P(A^c$\cap$B^c$\cap$C)

P(A^c$\cap$B^c) = 1 - P(A$\cup$B) = 1 - [P(A)+P(B)-P(A$\cap$B)] = 1 - [0.3+0.3] = 0.4

P(A^c$\cap$B^c$\cap$C) = P(A^c$\cap$B^c) - P(A^c$\cap$B^c$\cap$C^c) = 0.4-0.1 = 0.3