Question

Suppose that A and B are events such that $P\left(A\right)=\frac{1}{4}$,$P\left(B\right)=\frac{1}{3}$and $P(AorB)=\frac{1}{2}$

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

Part (b) Find$P(A\&B)$

Short answer

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

Part (b)$P(A\&B)=\frac{1}{12}$

Step-by-step solution

Part (a) Step 1. Given information.  

$$\begin{gathered} P\left(A\right)=\frac{1}{4} \\ P\left(B\right)=\frac{1}{3} \\ P(A\&B)=\frac{1}{2} \end{gathered}$$

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

$$\begin{gathered} P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B) \\ \frac{1}{2}=\frac{1}{4}+\frac{1}{3}-P(A\cap B) \\ P(A\cap B)=\frac{7}{12}-\frac{1}{2} \\ P(A\cap B)=\frac{1}{12} \end{gathered}$$

$A$and $B$are not mutually exclusive because $A$ and $B$ are not zero.

Part (b) Step 1. Given information.  

$$\begin{gathered} P\left(A\right)=\frac{1}{4} \\ P\left(B\right)=\frac{1}{3} \\ P(A\&B)=\frac{1}{2} \end{gathered}$$

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

$$\begin{gathered} P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B) \\ \frac{1}{2}=\frac{1}{4}+\frac{1}{3}-P(A\cap B) \\ P(A\cap B)=\frac{7}{12}-\frac{1}{2} \\ P(A\cap B)=\frac{1}{12} \end{gathered}$$