The State Police are trying to crack down on speeding on a particular portion of

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 97% of the time; that is, there is a 0.97 probability that the gun will detect a speeder, if the driver is actually speeding. Assume there is a 2% chance that the gun erroneously detects a speeder even when the driver is below the speed limit. Suppose that 94% of the drivers drive below the speed limit on this stretch of the Massachusetts Turnpike.

What is the probability that the gun detects speeding and the driver was speeding? (Round your answer to 4 decimal places.)

What is the probability that the gun detects speeding and the driver was not speeding? (Round your answer to 4 decimal places.)

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.)

What is the probability that the gun detects speeding and the driver was speeding? (Round your answer to 4 decimal places.)

Explanation / Answer

a. P(Detected and speeding) = P(Detected | Speeding) * P(Speeding)

= 0.97* (1-0.94) = 0.0582

b. P(Detected an not speeding) = P(Detected | Not Speeding) * P(Not speeding)

= 0.02 * 0.94 = 0.0188

c. By theorem of total probability,

P(Detected) = P(Detected and speeding) + P(Detected an not speeding)

= 0.0582 + 0.0188

= 0.077

By Bayes' theorem,

P(Speeding | Detected) = P(Detected | Speeding) * P(Speeding) / P(Detected)

= P(Detected and speeding) / P(Detected) = 0.0582 / 0.0770

= 0.7558

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