4 Suppose police in Allston are stopping random drivers on a

4. Suppose police in Allston are stopping random drivers on a Friday night and testing for DUIs, that the probability of a randomly-selected driver actually being drunk is 1/20 (hey, it\'s Friday night in Allston!), and the probability of a \'\'false positive\'\' in the Breathalyzer test is 1%. If you are stopped and test positive, what is the probability that you are actually drunk?

Solution

4.

Here,

P(positive) = P(drunk) P(positive|drunk) + P(not drunk) P(positive|not drunk)
= (1/20) (1-0.01) + (19/20)(0.01)

= 0.059

Now,

P(positive and drunk) = P(drunk) P(positive|drunk) = (1/20) (1-0.01)

= 0.0495

Hence,

P(drunk|positive) = P( positive and drunk) / P(positive) = 0.0495/0.059

= 0.83898 [ANSWER]