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 = <u><em>event that the police stops the driver</em></u>
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)
P(S/D) + P(B)
P(S/B) + P(G)
P(S/G)
= (0.08
0.85) + (0.32
0.38) + (0.60
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) = ![\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)}](https://tex.z-dn.net/?f=%5Cfrac%7BP%28D%29%20%5Ctimes%20P%28S%2FD%29%7D%7BP%28D%29%20%5Ctimes%20P%28S%2FD%29%2BP%28B%29%20%5Ctimes%20P%28S%2FB%29%2BP%28G%29%20%5Ctimes%20P%28S%2FG%29%7D)
= ![\frac{0.08\times 0.85}{0.08\times 0.85+0.32\times 0.38+0.60\times 0.02}](https://tex.z-dn.net/?f=%5Cfrac%7B0.08%5Ctimes%200.85%7D%7B0.08%5Ctimes%200.85%2B0.32%5Ctimes%200.38%2B0.60%5Ctimes%200.02%7D)
=
= 0.337
Hence, the probability that a driver who is stopped at the checkpoint is drunk is 0.337.