Question

Given that $P(A)=0.1,P(B|A)=0.6$ and $P(B|A^c)=0.3$ what is $P(A|B)$ ?

• A. $\frac{2}{11}$
• B. $\frac{4}{11}$
• C. $\frac{7}{11}$
• D. $\frac{9}{11}$

Answer 1

Answer: (A)

Given, $P(B|A)=0.6$
$\Rightarrow \frac{P\left(B\cap A\right)}{P\left(A\right)}=0.6$
$\Rightarrow P(B\cap A)=0.6\times 0.1=0.06$
Also, $P(B|A^C)=0.3$
$\Rightarrow \frac{P\left(B\cap A^C\right)}{P\left(A^C\right)}=0.3$
$\Rightarrow \frac{P\left(B\right)-P\left(B\cap A\right)}{1-P\left(A\right)}=0.3$
$\Rightarrow \frac{P\left(B\right)-0.06}{0.9}=0.3$
$\Rightarrow P\left(B\right)=0.33$
Now, $P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{0.06}{0.33}=\frac{2}{11}$