Question

Two events, A and B, are independent, with$\mathit{P}\left(A\right)\mathbf{=}\mathbf{.}\mathbf{3}$and$\mathit{P}\left(B\right)\mathbf{=}\mathbf{.}\mathbf{1}$.

a.Are A and B mutually exclusive? Why?

b.Find$\mathit{P}\left(A/B\right)$and$\mathit{P}\mathbf{(}\mathbf{B}\mathbf{/}\mathbf{A}\mathbf{)}$.

c.Find$\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{)}$.

Short answer

a. No, event A and B are not mutually exclusive.

b.The values are$P\left(A/B\right)=0.3$ and $P\left(B/A\right)=0.1$.

c. The value of$P\left(A\cup B\right)$ is 0.37.

Step-by-step solution

Definitions

The events are said to be mutually excusive that event that cannot occurs at the same time and having probability is zero.

The events are said independent if their probability does not affect each another.$\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right|}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\left)}\mathbf{ORP}\mathbf{\right(}\mathbf{B}\mathbf{\left|}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}$. And$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}$

The require formula for union is$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)=}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{)+}\mathbf{P}\mathbf{(}\mathbf{B}\mathbf{)-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}$.

Find A and B mutually exclusive

a.

Since the events are independent $P\left(A\right|B)\ne P(A)$.

Thus, event A and B are not mutually exclusive.

Find P(A/B) and P(B|A).

b.

$$\begin{gathered} P\left(A\backslash B\right)=\frac{P\left(A⌒B\right)}{P\left(B\right)} \\ =\frac{P\left(A\right).P\left(B\right)}{P\left(B\right)} \\ =0.3 \\ P\left(B\backslash A\right)=\frac{P\left(A⌒B\right)}{P\left(B\right)} \\ =\frac{P\left(A\right).P\left(B\right)}{P\left(A\right)} \\ =0.3 \end{gathered}$$

So, the values are$P\left(A\backslash B\right)=0.3$ and$P\left(B\backslash A\right)=0.1$

Find 

c.

Here,

$$\begin{gathered} P\left(A\cap B\right)=P\left(A\right).P\left(B\right) \\ =\left(0.3\right)\left(0.1\right) \\ =0.03 \end{gathered}$$

$$\begin{gathered} P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right) \\ =0.3+0.1-0.03 \\ =0.37 \end{gathered}$$

Therefore, the value of $P\left(A\cup B\right)$is 0.37.