Question

If A and B are two events such that

$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3}$ and $P(A\cap B)=\frac{1}{4},$ then find

(i) $P\left(\frac{A}{B}\right).$

(ii) $P\left(\frac{A^{\prime }}{B}\right).$

(iii) $P\left(\frac{A^{\prime }}{B^{\prime }}\right)$

Answer 1

Here, $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3}$ and $P(A\cap B)=\frac{1}{4}$

(i) $P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4}$

(ii) $P\left(\frac{A^{\prime }}{B}\right)=1-P\left(\frac{A}{B}\right)=1-\frac{3}{4}=\frac{1}{4}orP\left(\frac{A^{\prime }}{B}\right)=\frac{P(A^{\prime }\cap B)}{P\left(B\right)}=\frac{P\left(B\right)-P(A\cap B)}{P\left(B\right)}=\frac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{3}}=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}$

(iii) $P\left(\frac{A^{\prime }}{B^{\prime }}\right)=\frac{P(A^{\prime }\cap B^{\prime })}{P\left(B^{\prime }\right)}=\frac{1-P(A\cup B)}{1-P\left(B\right)}=\frac{1-\left[P\right(A)+P(B)-P(A\cap B\left)\right]}{1-P\left(B\right)}=\frac{1-[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}]}{1-\frac{1}{3}}=\frac{1-(\frac{5}{6}-\frac{1}{4})}{\frac{2}{3}}=\frac{1-\frac{14}{24}}{\frac{2}{3}}=\frac{\frac{10}{24}}{\frac{2}{3}}=\frac{30}{48}=\frac{5}{8}$