Question

$P\left(A\right)=\frac{3}{8};P\left(B\right)=\frac{1}{2};P(A\cup B)=\frac{5}{8}$, which of the following do/does hold good?

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

Answer 1

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

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

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
or $\frac{5}{8}=\frac{3}{8}+\frac{4}{8}-P(A\cap B)$
or $P(A\cap B)=\frac{2}{8}=\frac{1}{4}$
Now,
$P\left(A^c/B\right)=\frac{P(A^c\cap B)}{P\left(B\right)}$
$=\frac{P\left(B\right)-P(A\cap B)}{P\left(B\right)}$
$=1-2\left(\frac{1}{4}\right)$
$=\frac{1}{2}$
$2P\left(A/B^c\right)=\frac{2P(A\cap B^c)}{P\left(B^c\right)}$
$=\frac{2\left(P\right(A)-P(A\cap B\left)\right)}{P\left(B^c\right)}$
$=4(\frac{3}{8}-\frac{2}{8})=\frac{1}{2}$
Hence, option (a) is correct.
$P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
$=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}=P\left(B\right)$
Hence, (b) is correct. Again,
$P\left(A^c/B^c\right)=\frac{P(A^c\cap B^c)}{P\left(B^c\right)}$
$=\frac{1-P(A\cap B)}{1-P\left(B\right)}$
$=2(1-\frac{5}{8})=\frac{3}{4}$
$P\left(B/A^c\right)=\frac{P(B\cap A^c)}{1-P\left(A\right)}$
$=\frac{P\left(B\right)-P(A\cap B)}{\frac{5}{8}}$
$=\frac{(\frac{1}{2}-\frac{1}{4})}{\frac{5}{8}}=\frac{2}{5}$
Hence,
$8P\left(A^c/B^c\right)=15P\left(B/A^c\right)$
Hence, (c) is not correct. Again,
$2P\left(A/B^c\right)=\frac{1}{2}$
or $P\left(A/B^c\right)=\frac{1}{4}=P(A\cap B)$
Hence, (d) is correct