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On a Saturday night 20% of all drives in U.S.A. are under the influence of alcohol. The probability that a driver under the influence of alcohol will have an accident is 0.001. The probability that a sober driver will have an accident is 0.0001. If a car on a Saturday night smashed into a tree, the probability that the driver was under the influence of alcohol is

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$\frac{3}{7}$

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{6}{7}$

answer is C.

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Detailed Solution

A= Car met with an accident

$B_{1} :$ Driver was alcoholic $P (B_{1}) = \frac{1}{5}$

$B_{2} :$ Driver was sobed $P (B_{2}) = \frac{4}{5}$

$P (A (B_{1})) = 0.001, P (A (B_{2})) = 0.0001$

$Re q . = \frac{\frac{1}{5} . \frac{1}{1000}}{\frac{1}{5} . \frac{1}{1000} + \frac{4}{5} . \frac{1}{10000}} = \frac{5}{7}$

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