A group of policemen have breathalyzers displaying false drunkeness in 5% of the

Question

A group of policemen have breathalyzers displaying false drunkeness in 5% of the cases. However, the breathalyzers never fail to detect a truly drunk person. 1/1000 of drivers are driving drunk. Suppose the policemen then stop a driver at random, and force them to take a breathalyzer test. It indicates that he or she is drunk. We assume you don't know anything else about him or her. How high is the probability he or she really is drunk?

a. Greater than or equal to .2.

b. Less than .2 but more than .1.

c. Less than .1 but more than .05.

d. Less than or equal to .05.

e. There is not enough information to calculate the probability.

Explanation / Answer

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(1)

P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(2)

If A is made up of k mutually and collectively exhaustive sub-events, A1, A2,..Ak,

P(B) = sum over i = 1 to k of {P(B/Ai) x P(Ai)} ………………………………..(3)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..………………….(4)

Now, to work out the solution,

Let A represent the event the driver is drunk and B represent the event the breathalyzer test indicates that driver is drunk. Naturally then, AC represents the event the driver is not drunk and BC represents the event the breathalyzer test indicates that driver is not drunk.

We will now translate the given data into probabilities of A and B.

A group of policemen have breathalyzers displaying false drunkenness in 5% of the cases.

=> P(B/AC) = 0.05 ……………………………………………………………………(5)

The breathalyzers never fail to detect a truly drunk person. => P(B/A) = 1 …………(6)

1/1000 of drivers are driving drunk. => P(A) = 0.001……………………………….(7)

Suppose the policemen then stop a driver at random, and force them to take a breathalyzer test. It indicates that he or she is drunk. => B has occurred…………………………..(8)

We want the probability he or she really is drunk under (8), i.e., P(A/B) …………….(9).

Now, by (4), P(A/B) = P(B/A) x { P(A)/P(B)} = 1 x {0.001/P(B)} = {0.001/P(B)} …(10)

By (2), P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}

= (1 x 0.001) + (0.05 x 0.999) [P(AC) = 1 – P(A) = 1 – 0.001 from (7)]

= 0.05095 ……………………………………………………………………………..(11)

(10) and (11) => P(A/B) = 0.001/0.05095 = 0.0196 ANSWER Option (d)

DONE

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