The probability that Joe passes his theory is 0.9. If he passes his theory, the probability of passing his practical is 0.7; otherwise the chance of passing is only 0.4. Given that he fails his driving license, what is the probability that he fails because of his practical only?
Do everything step by step to avoid confusion.
1) P( pass theory) = 0.9 therefore P (not pass theory) = 0.1
2) P( pass prac) = 0.7 therefore P (not pass prac) = 0.3
3) P(not pass th & pass prac)= 0.4 therefore P(not pass th & not pass prac)= 0.6
`therefore P` (pass Th & Prac) `= (0.9 times 0.7) = 0.63`
As this represents him obtaining his licence,
`therefore P` (not pass th & prac) `= 0.37 (1-0.63)`
`therefore P` ( failed because of prac) `= 0.37/0.63`
The probability that he failed because of his practical = 0.59 or 59%
Another way to do this is by drawing a tree diagram.
We see that in 3 of the scenarios he fails. .27, .04, and .06 are the probability of each happening. The probability that he fails only because of the practical is .27.
Given that he fails, the probability that he failed only because of his practical is found by taking probability of him failing only his practical, and then dividing by all the probabilities of him failing in general.
`.27/(.27+.04+.06) = .73`
`P` (not obtaining licence) `= 0.37 (1-0.63)`
`therefore P` (pass th and not pass prac) `= ( 0.9 times 0.3)= 0.27` (see 1) and 2) above)
`therefore P` (failed because of prac) `= 0.27/0.37`
=0.73 or 73%
Ans: The probability that he failed because of his practical only is 0.73 or 73%