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  4. If P(A) = (1/4), P(B) = (1/5) and P(AB) = (1/8) then P((AC/BC)) =

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Q. 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) = $

J & K CETJ & K CET 2017Probability - Part 2

Solution:

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