Question

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

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

Answer 1

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

The correct options are
A $15P\left(A^C/B^C\right)=8P\left(B/A^C\right)$
B $P\left(B\right)=P\left(A^C/B\right)$
C $P\left(A/B^C\right)=(A\cap B)$
D $P\left(A^C/B\right)=2P\left(A/B^C\right)$

We have

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

Now,

$P\left(A^c/B\right)=\frac{P(A^c\cap B)}{P\left(B\right)}=\frac{P\left(B\right)-P(A\cap B)}{P\left(B\right)}=\frac{\frac{1}{2}-\frac{1}{4}}{\frac{1}{2}}=\frac{2}{4}=\frac{1}{2}=P\left(B\right)P\left(A/B^c\right)=\frac{P(A\cap B^c)}{P\left(B^c\right)}=\frac{P\left(A\right)-P(A\cap B)}{1-P\left(B\right)}=\frac{\frac{3}{8}-\frac{1}{4}}{1-\frac{1}{2}}=\frac{1}{4}=P(A\cap B)$

Hence,

$P\left(A^c/B\right)=2P\left(A/B^c\right)P\left(A^c/B^c\right)=\frac{P(A^c\cap B^c)}{P\left(B^c\right)}=\frac{1-P(A\cup B)}{1-P\left(B\right)}=\frac{1-\frac{5}{8}}{\frac{1}{2}}=\frac{3}{4}andP\left(B/A^c\right)=\frac{P(A^c\cap B)}{P\left(A^c\right)}=\frac{P\left(B\right)-P(A\cap B)}{1-P\left(A\right)}=\frac{\frac{1}{2}-\frac{1}{4}}{1-\frac{3}{8}}=\frac{2}{5}$

Thus we see that

$\frac{P\left(A^c/B^c\right)}{P\left(B/A^c\right)}=\frac{\frac{3}{4}}{\frac{2}{5}}=\frac{15}{8}\Rightarrow 8P\left(A^c/B^c\right)=15P\left(B/A^c\right)$