Question

3.   Suppose A and B are two events for which P(A) =0.18, P(B) = 0.45, and

        P(A or B) = 0.54.

1. Find P( A and B)

2. Find $P\left (A \mid B\right )$

3. Are A and B mutually exclusive?          (Support your answers!)

4. Are A and B independent?

Answer 1

Let A and B be two events with P(A)=0.18, and P(B)=0.45, and P(A or B)=0.54.

We making use of the theorem of probability based on addition which is given by

$P(A \bigcup B)=P(A)+P(B)-P(A \bigcap B). \\ where \ P(A \bigcup B) =P(A \ or \ B) \\ P(A \bigcap B)=P(A\ and \ B).$a. Therefore we have

$\\ Since \ P(A \bigcup B) =P(A \ or \ B)=0.54 \\ P(A )=0.18, P( B)=0.45.\ then \\ P(A \bigcup B)=P(A)+P(B)-P(A \bigcap B). \\ 0.54=0.18+0.45-P(A \bigcap B). \\ \Rightarrow P(A \bigcap B)=0.63-0.54= 0.09.$Therefore the probability of P(A and B)=0.09.

b.

$P(A|B)=\frac{P(A \bigcap B)}{P(B)}. \\ Since \ P( B)=0.45\ and \ P(A \bigcap B)= 0.09. \\ Therefore\ P(A|B)=\frac{0.09}{0.45}=0.2. \\ \Rightarrow P(A|B)=0.2.$c . Event A and B is said to be mutually exclusive if the occurrence of one exclude the occurrence of other. In other words events A and B is said to be mutually exclusive if

$P(A \bigcap B)=0.$

But we have

$P(A \bigcap B)=0.09.$

Therefore A and B is not mutually exclusive events.

d. Two events A and B is said to be independent if it satisfies

$P(A \bigcap B)=P(A)*P(B).$

We have

$P(A \bigcap B)=0.09 , \ P(A)=0.18, \ and \ P(B)=0.45.$Therefore we have

$\\ P(A)*P(B)=0.18*0.45=0.081. \\ Since \ P(A \bigcap B)=0.09, so \\ P(A)*P(B)\neq P(A \bigcap B).$

Therefore A and B is not independent.