Question

If $P\left(A\right)=\frac{6}{11},P\left(B\right)=\frac{5}{11}$ and $P(A\cup B)=\frac{7}{11}$, find
(i) $P(A\cap B)$
(ii) $P\left(A|B\right)$
(iii) $P\left(B|A\right)$

• A. $0.58,0.58,0.83$
• B. $0.42,0.67,0.57$
• C. $0.36,0.80,0.66$
• D. $0.80,0.98,0.87$

Answer 1

Answer: (C)

$0.36,0.80,0.66$

It is given that $P\left(A\right)=\frac{6}{11},P\left(B\right)=\frac{5}{11}$ and $P(A\cup B)=\frac{7}{11}$
(i) $P(A\cup B)=\frac{7}{11}$
$\therefore P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{7}{11}$
$\Rightarrow \frac{6}{11}+\frac{5}{11}-P(A\cap B)=\frac{7}{11}$
$\Rightarrow P(A\cap B)=\frac{11}{11}-\frac{7}{11}=\frac{4}{11}=0.36$
(ii) It is known that, $P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
$\Rightarrow P\left(A|B\right)=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}=0.80$
(iii) It is known that, $P\left(B|A\right)=\frac{P(A\cap B)}{P\left(A\right)}$
$\Rightarrow P\left(B|A\right)=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}=0.66$