Question

The State Police are trying to crack down on speeding on a particular portion of the Massachusetts Turnpike. To aid in this pursuit, they have purchased a new radar gun that promises greater consistency and reliability. Specifically, the gun advertises ± one-mile-per-hour accuracy 98% of the time; that is, there is a 0.98 probability that the gun will detect a speeder, if the driver is actually speeding. Assume there is a 1% chance that the gun erroneously detects a speeder even when the driver is below the speed limit. Suppose that 95% of the drivers drive below the speed limit on this stretch of the Massachusetts Turnpike. a. What is the probability that the gun detects speeding and the driver was speeding? (Round your answer to 4 decimal places.) b. What is the probability that the gun detects speeding and the driver was not speeding? (Round your answer to 4 decimal places.) c. Suppose the police stop a driver because the gun detects speeding. What is the probability that the driver was actually driving below the speed limit? (Round your answer to 4 decimal places.)

Answer 1

Explanation:

We define 3 events

A = event that driver is above speed limit

B = event that driver is below limit

T = event that the gun detected him

At 95%

P(A) = 1-0.95 = 0.05

P(B) = 0.95

P(T|A) = 0.98

P(A|T) = 0.02

P(T|B) = 1-0.99 = 0.01

P(B|T) = 0.99

1. For answer a

P(TnA) = P(A) x P(T|A)

= 0.05 x 0.98

= 0.049 is the probability gun detected speeding and the driver was speeding

2. For answer b

P(TnB) = P(B) x P(T|B)

= 0.95x0.01

= 0.0095 is probability that gun deects speeding and driver was not speeding

3. For answer c

We solve this using bayes theorem

P(B|T) = P(B) x P(T|B) / (P(B)*P(T|B)) + (P(A) +P(T|A))

= 0.095x0.01 divided by (0.95x0.01)+(0.05*0.98)

= 0.0095 divided by 0.0585

= 0.16239

= 0.1624 to 4 decimal places.