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, $P\left(A | B\right)=\frac{P\left(A \,\cap \,B\right)}{P\left(B\right)}=\frac{0.06}{0.33}=\frac{2}{11}$