Question

If $P\left(A\right)=\frac{1}{4},P\left(B\right)=\frac{1}{5}$ and $P\left(AB\right)=\frac{1}{8}$ then $P\left(\frac{A^C}{B^C}\right)=$

• A. $\frac{21}{32}$
• B. $\frac{25}{32}$
• C. $\frac{27}{32}$
• D. $\frac{29}{32}$

Answer 1

Answer: (C)

We get , $P\left(A\right)=\frac{1}{4}$,
$P\left(B\right)=\frac{1}{5}$
and $P\left(AB\right)=\frac{1}{8}$
$P\left(\frac{A^C}{B^C}\right)$
$=\frac{P\left(A^C\cap B^C\right)}{P\left(B^C\right)}$
$\frac{P\left({\left(A\cup B\right)}^C\right)}{P\left(B^C\right)}$
$=\frac{1-P\left(A\cup B\right)}{1-P\left(B\right)}$
$=\frac{1-\left[P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\right]}{1-P\left(B\right)}$
$=\frac{1-\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}}$
$=\frac{40-10-8+5}{40\times \left(\frac{4}{5}\right)}$
$=\frac{27}{32}$