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On Saturday night 20% of all drivers in 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, 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

(3)
A: car met with an accident
E₁: driver was alcoholic, P(E₁) = 1/5
E₂: driver was sober, P(E₂) = 4/5
P(A/E₁) = 0.001, P(A/E₂) = 0.0001
$P (E_{1} / A) = \frac{(0.2) (0.001)}{(0.2) (0.001) + (0.8) (0.0001)} = 5 / 7$

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