Mathematics - Probability Question with Solution | TestHub

Given,

 $A$ and $B$ be two events such that $P\left(B\mid A\right)=\frac{2}{5}$, $P\left(A\mid B\right)=\frac{1}{7}$ and $P\left(A\cap B\right)=\frac{1}{9}$, 

Now using the formula of conditional probability we get,

$P\left(A\mid B\right)=\frac{1}{7}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{7}$

$\Rightarrow \frac{{\displaystyle \frac{1}{9}}}{P\left(B\right)}=\frac{1}{7}$

$\Rightarrow P\left(B\right)=\frac{7}{9}$

Now again using formula we get,

$P\left(B\mid A\right)=\frac{2}{5}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{2}{5}$

$\Rightarrow \frac{{\displaystyle \frac{1}{9}}}{P\left(A\right)}=\frac{2}{5}$

$\Rightarrow P\left(A\right)=\frac{5}{18}$

Now we know that,

 $P\left(A^{\text{'}}\cup B\right)=1-P\left(A\cup B\right)+P\left(B\right)$

$\Rightarrow P\left(A^{\text{'}}\cup B\right)=1-P\left(A\right)+P\left(A\cap B\right)=\frac{5}{6}$

And $P\left(A^{\text{'}}\cap B^{\text{'}}\right)=1-P\left(A\cup B\right)$ 

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

$\Rightarrow P\left(A^{\text{'}}\cap B^{\text{'}}\right)=1-P\left(A\right)-P\left(B\right)+P\left(A\cap B\right)=\frac{1}{18}$

$\Rightarrow$ Both $\left(S1\right)$ and $\left(S2\right)$ are true.