On a Saturday night, 20% of all drivers in the USA 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, then the probability that the driver was under the influence of alcohol, is

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$\frac{5}{7}$

$\frac{3}{7}$

$\frac{4}{7}$

$\frac{6}{7}$

answer is C.

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Detailed Solution

Let $E_{1}, E_{2}$ and A be the event defined as follow:

$E_{1}$ : The driver are not under the influence

$E_{2} :$ The driver are under the influence of alcohol

A : Car met with an accident

Now,

$\begin{array}{l} P (E_{1}) = \frac{80}{100} = \frac{4}{5} \\ P (E_{2}) = \frac{20}{100} = \frac{1}{5} \\ P (\frac{A}{E_{1}}) = 0.0001 \\ P (\frac{A}{E_{2}}) = 0.001 \end{array}$

To find, $P (\frac{E_{2}}{A})$

By Baye' s theorem, we have

$\begin{array}{l} P (\frac{E_{2}}{A}) = \frac{P (E_{2}) P (\frac{A}{E_{2}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}})} = \frac{\frac{1}{5} \times 0.001}{\frac{4}{5} \times 0.0001 + \frac{1}{5} \times 0.001} \\ = \frac{0.001}{0.0004 + 0.001} = \frac{0.001}{0.0014} = \frac{10}{14} = \frac{5}{7} \end{array}$