Question

Given that the event A and B ar esuch that $P\left(A\right)=\frac{1}{2}P(A\cup B)=\frac{3}{5}andP\left(B\right)=p.$ Find p, if they are
mutually exclusive
independent

Answer 1

When A and B are mutually exclusive, then $A\cap B=\varphi \Rightarrow P(A\cap B)=0\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\Rightarrow \frac{3}{5}=\frac{1}{2}+p-0\Rightarrow p=\frac{3}{5}-\frac{1}{2}=\frac{6-5}{10}=\frac{1}{10}$

When A and B are independent events, then
$P(A\cap B)=P\left(A\right).P\left(B\right)=\frac{1}{2}P(\because P(A)=\frac{1}{2}P(B)=Pgiven)Now,P(A\cap B)=P\left(A\right)P\left(B\right)=\frac{1}{2}P\Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{1}{2}p\Rightarrow \frac{3}{5}-\frac{1}{2}=\frac{2p-p}{2}\frac{6-5}{10}\Rightarrow P=\frac{2}{10}=\frac{1}{5}$