Question

Let $A$ & $B$ be two independent events such that $P\left(A\right)=1/3$ & $P\left(B\right)=1/4$ match the entries from the following two column

Answer 1

$P\left(A\right)=1/3,P\left(B\right)=1/4$

$\because$ the events are independent, so,

$P\left(A\right)\cap P\left(B\right)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{12}$

Now,

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=1/3+1/4-1/2$

$=\frac{7}{12}-\frac{1}{12}=\frac{6}{12}=1/2$

$\Rightarrow P(A\cup B)=1/\mathrm{2...}\left(1\right)$

& $P\left(\frac{A}{A\cup B}\right)=\frac{1}{12}...\left(2\right)$

& from (1), $P(A\cup B)=1-P(A^{\prime }\cap B^{\prime })$

$P(A^{\prime }\cap B^{\prime })=1/2$

$\Rightarrow P\left(\frac{A^{\prime }}{B^{\prime }}\right)=\frac{P(A^{\prime }\cap B^{\prime })}{P\left(B^{\prime }\right)}=\frac{1/2}{1-P\left(B\right)}=\frac{1/2}{3/4}=\frac{2}{3}$

$\Rightarrow P\left(\frac{A^{\prime }}{B^{\prime }}\right)=2/\mathrm{3...}\left(3\right)$