Question

4. Suppose the probability of event A is 0.23, the probability of the intersection of event A and event Bis 0.12, and

the probability of the union of event A and event B is 0.34. Find the probability of the complement of event B.
(A) 0.11
(B) 0.23
(C) 0.48
(D) 0.52
(E) 0.77

Answer 1

`P(B') = 0.77`

Explanation:

Given

`P(A) = 0.23`

`P(A\ n\ B) = 0.12`

`P(A\ u\ B) = 0.34`

Required

Find P(B')

First, we calculate P(B).

We have:

`P(A\ n\ B) = P(A) + P(B) - P(A\ u\ B)`

Substitute values

`0.12 = 0.23 + P(B) - 0.34`

`0.12 = 0.23 - 0.34 + P(B)`

Collect like terms

`P(B) = 0.12 - 0.23 + 0.34`

`P(B) = 0.23`

P(B') is then calculated as:

`P(B') = 1 - P(B)`

`P(B') = 1 -0.23`

`P(B') = 0.77`