Question

Let $A$ and $B$ be two events such that $P\left(A\right)=\frac{3}{8},P\left(B\right)=\frac{1}{2}$ and $P(A\cup B)=\frac{5}{8}$. Then which of the following do/does hold good?

• A. $P\left(A^C|B\right)=2P\left(A|B^C\right)$
• B. $P\left(B\right)=P\left(A|B\right)$
• C. $15P\left(A^C|B^C\right)=8P\left(B|A^C\right)$
• D. $P\left(A|B^C\right)=P(A\cap B)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $P\left(A^C|B\right)=2P\left(A|B^C\right)$
B $P\left(B\right)=P\left(A|B\right)$
D $P\left(A|B^C\right)=P(A\cap B)$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow \frac{5}{8}=\frac{3}{8}+\frac{1}{2}-P(A\cap B)$
$\Rightarrow P(A\cap B)=\frac{2}{8}=\frac{1}{4}$
Now, $P\left(A^C|B\right)=\frac{P(A^C\cap B)}{P\left(B\right)}$
$=\frac{P\left(B\right)-P(A\cap B)}{P\left(B\right)}$
$=1-2\left(\frac{1}{4}\right)$
$=\frac{1}{2}$
and $2P\left(A|B^C\right)=\frac{2P(A\cap B^C)}{P\left(B^C\right)}$
$=\frac{2(P\left(A\right)-P(A\cap B))}{1-P\left(B\right)}$
$=4(\frac{3}{8}-\frac{1}{4})=\frac{1}{2}$
$P\left(A^C|B\right)=2P\left(A|B^C\right)$

$P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}=P\left(B\right)$

Again, $P\left(A^C|B^C\right)=\frac{P(A^C\cap B^C)}{P\left(B^C\right)}$
$=\frac{1-P(A\cup B)}{1-P\left(B\right)}$
$=2(1-\frac{5}{8})=\frac{3}{4}$
and $P\left(B|A^C\right)=\frac{P(B\cap A^C)}{1-P\left(A\right)}$
$=\frac{P\left(B\right)-P(A\cap B)}{5/8}$
$=\frac{1/2-1/4}{5/8}$
$=\frac{1}{4}\times \frac{8}{5}$
$=\frac{2}{5}$
Hence, $8P\left(A^C|B^C\right)=15P\left(B|A^C\right)$

Again, $2P\left(A|B^C\right)=\frac{1}{2}$
$\Rightarrow P\left(A|B^C\right)=\frac{1}{4}=P(A\cap B)$