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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 75% of the time; that is, there is a 0.75 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 72% 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?
b. What is the probability that the gun detects speeding and the driver was not speeding?
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?

Respuesta :

Carincon97

Answer:

a) P(A∩B) = 0.21

b) P(A∩B') = 0.0072

c) P(B'|A)=0.0072/0.2172=0.0331

Step-by-step explanation:

A =  the gun will detect a speeder

B =  driver is actually speeding

P(A|B) = 0.75

P(A|B')=0.01

P(B') = 0.72

a) by definition

P(A∩B)=P(A|B)*P(B)=0.75*(1-0.72)=0.21

b) by definition

P(A∩B')=P(A|B')*P(B')=0.01*(0.72)=0.0072

c)  

by bayes theorem

P(B'|A)=P(A|B')*P(B')/P(A)

by total probability theorem

P(A)=P(A∩B)+P(A∩B')=0.21+0.0072=0.2172

so

P(B'|A)=0.0072/0.2172=0.0331

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