A police checkpoint is set up on New Years Eve to catch drunk drivers. It is known that 8% of drivers on New Years Eve have a blood-alcohol level that is above the legal limit. It is also known that 32% of drivers in the area never drink and drive, but are simply bad drivers. The remaining 60% of drivers never drink and drive and are good drivers. At this particular checkpoint, the police stop 85% of the drunk drivers, stop 38% of the bad drivers, and stop 2% of the good drivers. (a) There are four key events in this problem. Specify the four key events and define notation for each of them. (b) What is the probability that a randomly selected driver will be stopped at the checkpoint? (c) What is the probability that a driver who is stopped at the checkpoint is drunk?

Question

30-09-2020
Mathematics

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A police checkpoint is set up on New Years Eve to catch drunk drivers. It is known that 8% of drivers on New Years Eve have a blood-alcohol level that is above the legal limit. It is also known that 32% of drivers in the area never drink and drive, but are simply bad drivers. The remaining 60% of drivers never drink and drive and are good drivers. At this particular checkpoint, the police stop 85% of the drunk drivers, stop 38% of the bad drivers, and stop 2% of the good drivers. (a) There are four key events in this problem. Specify the four key events and define notation for each of them. (b) What is the probability that a randomly selected driver will be stopped at the checkpoint? (c) What is the probability that a driver who is stopped at the checkpoint is drunk?

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Answer:

(a) Let the probability drivers are drunk = P(D) = 0.08

The probability that drivers in the area never drink and drive, but are simply bad drivers = P(B) = 0.32

The probability that drivers in the area never drink and drive, and are good drivers = P(G) = 0.60

Let S = event that the police stops the driver.

(b) The probability that a randomly selected driver will be stopped at the checkpoint is 0.2016.

(c) The probability that a driver who is stopped at the checkpoint is drunk is 0.337.

Step-by-step explanation:

We are given that 8% of drivers on New Years' Eve have a blood-alcohol level that is above the legal limit. It is also known that 32% of drivers in the area never drink and drive, but are simply bad drivers. The remaining 60% of drivers never drink and drive and are good drivers.

At this particular checkpoint, the police stop 85% of the drunk drivers, stop 38% of the bad drivers, and stop 2% of the good drivers.

(a) Let the probability drivers are drunk = P(D) = 0.08

The probability that drivers in the area never drink and drive, but are simply bad drivers = P(B) = 0.32

The probability that drivers in the area never drink and drive, and are good drivers = P(G) = 0.60

Let S = event that the police stops the driver

These are the four events stated in the question.

(b) So, the probability that the police stop the drunk drivers = P(S/D) = 0.85

The probability that the police stop the bad drivers = P(S/B) = 0.38

The probability that the police stop the good drivers = P(S/G) = 0.02

Now, the probability that a randomly selected driver will be stopped at the checkpoint is given by = P(S)

P(S) = P(D) [tex]\times[/tex] P(S/D) + P(B) [tex]\times[/tex] P(S/B) + P(G) [tex]\times[/tex] P(S/G)

= (0.08 [tex]\times[/tex] 0.85) + (0.32 [tex]\times[/tex] 0.38) + (0.60 [tex]\times[/tex] 0.02)

= 0.068 + 0.1216 + 0.012

= 0.2016

Hence, the probability that a randomly selected driver will be stopped at the checkpoint is 0.2016.

(c) Now, the probability that a driver who is stopped at the checkpoint is drunk is given by = P(D/S)

P(S/W) = [tex]\frac{P(D) \times P(S/D)}{P(D) \times P(S/D)+P(B) \times P(S/B)+P(G) \times P(S/G)}[/tex]

= [tex]\frac{0.08\times 0.85}{0.08\times 0.85+0.32\times 0.38+0.60\times 0.02}[/tex]

= [tex]\frac{0.068}{0.2016}[/tex] = 0.337

Hence, the probability that a driver who is stopped at the checkpoint is drunk is 0.337.