A breath analyzer, used by the police to test whether drivers exceed the legal l

Question

A breath analyzer, used by the police to test whether drivers exceed the legal limit set for the blood alcohol percentage while driving, is known to satisfy P(A|B) = P(A^c|B^c) = p where A is the event "breath analyzer indicates that legal limit is exceeded" and B" drivers blood alcohol percentage exceeds legal limit." On Saturday night about 5% of the drivers are known to exceed the limit. Describe in words the meaning of P (B^c|A). Determine P (B^c|A) if p = 0.95. Are the two events A and B independent? Should we expect them to be?

Explanation / Answer

Given,

Event A: breath analyser indicates that legal limit is exceeded

Event B: drivers blood alcohol percentage exceeds legal limit

5% of drivers are known to exceed the limit

P(B|A) = P(Bc|Ac) = p

A.a) P(Bc |A) means the probability that a driver who is shown as exceeding the legal limit actually did not exceed the legal limit.

A.b) P(Bc |A) if p=0.95

P(Bc |A)= P(P(Bc A) /P(A)

P(Bc |A)=P(A|Bc )*P(Bc ) /P(A)

P(Bc |A)= P(A|Bc )*P(Bc ) /P(A|Bc )*P(Bc ) + P(A|B)*P(B)

we know P(B)=0.05 and P(Bc )=0.95 (5% of drivers are known to exceeded the limit)

P(A|Bc )=1-P(Ac|Bc)=1-p    

we know P(A|B) = P(Bc|Ac) = p

P(Bc |A)= (1-p)*0.95/(1-p)*0.95+p*0.05

p is given as 0.95

solving we get,

P(Bc |A)= (1-0.95)*0.95/(1-0.95)*0.95+0.95*0.05

P(Bc |A)=0.0475/0.095

P(Bc |A)=0.5 or 50%

A.c) yes both the events are independent

event A is the test result of the breath analyser

Event B is the actual position of the driver

both are not dependent on each other

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