Question

Suppose that A and B are events such that $P\left(A\right)=\frac{1}{3}$, $P(AorB)=\frac{1}{2}$and role="math" localid="1651219326316" $P(A\&B)=\frac{1}{10}$

Part (a). Are event A and B mutually exclusive ? Explain your answer.

Part (b). Find role="math" localid="1651219700163" $P\left(B\right)$

Short answer

Part (a) $A$and $B$are not mutually exclusive events.

Part (b).$P(A\cup B)=\frac{11}{15}$

Step-by-step solution

Part (a) Step 1. Given information.  

$$\begin{gathered} P\left(A\right)=\frac{1}{3} \\ P(AorB)=\frac{1}{2} \\ P(A\&B)=\frac{1}{10} \end{gathered}$$

Part (a) Step 2.If the occurrence of one of two events A and B eliminates the occurrence of the other, they are mutually exclusive. A and B are not connected in this situation. 

$P(A\cap B)=\frac{1}{10}\ne 0$

This implies that certain outcomes are shared by both events A and B. As a result, A and B are not mutually exclusive.

Part (b) Step 1. Given information.  

$$\begin{gathered} P\left(A\right)=\frac{1}{3} \\ P(AorB)=\frac{1}{2} \\ P(A\&B)=\frac{1}{10} \end{gathered}$$

Part (b) Step 2. The following concept was used:

$$\begin{gathered} P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B) \\ P(A\cup B)=\frac{1}{3}+\frac{1}{2}-\frac{1}{10} \\ P(A\cup B)=\frac{10+15-3}{30} \\ P(A\cup B)=\frac{22}{30} \\ P(A\cup B)=\frac{11}{15} \end{gathered}$$