Mathematics - Probability Question with Solution | TestHub

$P\left(A\cap B\right)=\frac{1}{6}$ and $P\left(A^C\cap B^C\right)=\frac{1}{3}$

Now, $P{\left(A\cup B\right)}^C=P\left(A^C\cap B^C\right)=\frac{1}{3}$

$\Rightarrow 1-P\left(A\cup B\right)=\frac{1}{3}$

$\Rightarrow P\left(A\cup B\right)=\frac{2}{3}$

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

$\Rightarrow P\left(A\right)+P\left(B\right)=\frac{5}{6} ...\left(\mathrm{i}\right)$

$\because A$ and $B$ are independent events

$\therefore P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

$\Rightarrow P\left(A\right)P\left(B\right)=\frac{1}{6}$

${\left[P\left(A\right)-P\left(B\right)\right]}^2={\left[P\left(A\right)+P\left(B\right)\right]}^2-4P\left(A\right)P\left(B\right)$

$=\frac{25}{36}-\frac{4}{6}=\frac{1}{36}$

$\Rightarrow P\left(A\right)-P\left(B\right)=\pm \frac{1}{6} ...\left(\mathrm{ii}\right)$

On solving Eqs. (i) and (ii), we get

$P\left(A\right)=\frac{1}{2}\text{ or }\frac{1}{3}$

Hence, option (b) is correct.