Police are trying to crack down on speeding on a particular section of a highway
ID: 3341177 • Letter: P
Question
Police are trying to crack down on speeding on a particular section of a highway. To aid in this pursuit, they have purchased a new radar gun that promises greater consistency and reliability. Specifically, the gun will detect a speeder 98 percent of the time if the driver is speeding. The gun may also detect a driver as a speeder 1 percent of the time when the driver is not speeding. Suppose that 95 percent of the drivers drive below the speed limit on this section of the highway.
I’ve labeled the probabilities as the following (PLEASE CHECK LOGIC):
A = Driver is Speeding
B = Radar Detects Speeding Vehicle
A’ = Driver is NOT Speeding
B’ = Radar Does NOT Detect Speeding Vehicle
Then:
P(A) = .05 P(A’) = .95
P(B | A) = .98 P(B’ | A) = .02
P(B | A’) = .01 P(B’ | A’) = .99
1) What is the probability that a driver on this highway will be a speeder and will not be detected by the radar gun? SHOW ALL WORK using above Probability Labels.
2) 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? SHOW ALL WORK using above Probability Labels.
3) What is the probability that a driver will be detected by the radar gun on this highway? Briefly explain/ SHOW ALL WORK using above Probability Labels.
4) What is the probability that a driver will be neither a speeder nor will be detected by the radar gun on this highway? SHOW ALL WORK using above Probability Labels.
Explanation / Answer
1)probability that a driver on this highway will be a speeder and will not be detected by the radar gun
=P(A n Bc)=P(A)*P(Bc |A) =0.05*0.02=0.001
2)probability radar detects speeding =P(B)=P(A)*P(B|A)+P(Ac)*P(B|Ac)=0.05*0.98+0.95*0.01=0.0585
therefore probability that the driver was actually driving below the speed limit given gun detects speeding
=P(Ac|B) =P(Ac)*P(B|Ac)/P(B) =0.95*0.01/0.0585 =0.1624
3)
probability that a driver will be detected by the radar gun on this highway =P(B)=P(A)*P(B|A)+P(Ac)*P(B|Ac)=0.05*0.98+0.95*0.01=0.0585
4) probability that a driver will be neither a speeder nor will be detected by the radar gun on this highway
P(Ac nBc) = P(Ac) *P(Bc|Ac)=0.95*0.99 =0.9405
please revert for any clarificaiton required