# Given that the events A and B are such that P(A)

Question:

Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $P(B)=p$. Find $p$ if they are (i) mutually exclusive (ii) independent.

Solution:

It is given that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{3}{5}$, and $\mathrm{P}(\mathrm{B})=p$

(i) When $A$ and $B$ are mutually exclusive, $A \cap B=\Phi$

$\therefore P(A \cap B)=0$

It is known that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-0$

$\Rightarrow p=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

(ii) When $\mathrm{A}$ and $\mathrm{B}$ are independent, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} p$

It is known that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{1}{2} p$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+\frac{p}{2}$

$\Rightarrow \frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

$\Rightarrow p=\frac{2}{10}=\frac{1}{5}$