Need some help with this conditional probability question.
Jo is taking his driving test. There are 2 parts to his test: theory and practical. He needs to pass both to get his license. The probability that he passes his theory part is 0.8 and the probability that he passes the practical part once he has passed the theory part is 0.6. The probability that he passes the practical part if he failed the theory part is only 0.2.
You can draw a tree diagram for this.
Question: Given that he fails to get his license, find the probability that he only failed the practical part.
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The possibilities are:
(1) Jo fails the theory part
(2) Jo passes the theory, but fails the practical
(3) Jo passes both
(A) The probability of (1) is .2
(B) The probability of (2) is (.8)(.4)=.32
(C) The probability of (3) is (.8)(.6)=.48
Note that the probabilites sum to 1 as expected.
The probability that Jo only failed the practical part is .32
** Let PT=pass theory and FP= failed practical. Then P(FP)=P(FP|PT)P(PT) or P(FP)=(.4)(.8)=.32
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