Homework Question:
The probability that a person passes a test on the first try is $0.65$.
The probability that a person who fails the first test on the second try is $0.75$.
The probabilty that a person who fails the first and the second test will pass the third time is $0.60$.
Find the probability that a person fails the first and second tests and passes on third try.
My Solution:
P(P1) = passing on the 1st attempt = $0.67$
P(P2|F1) = passing on the 2nd attempt given that you fail the 1st attempt = $0.78$
P(P3|F2) = $0.63$
P(F1) = 1-P(P1) = $0.33$
P(F2|F1) = 1-P(P2|F1) = $0.22$
P(F3|F2) = 1-P(P3|F2) = $0.37$
The answer I have is P(F1)*P(F2)*P(P3) = $0.33 \cdot 0.22 \cdot 0.37 = 0.0525$.
I'm not sure if I'm on the right track here. Any help would be appreciated.
Best Answer
You are on the right track but have derailed somewhere.
Where did you get the numbers in your attempted solution? They are different from the ones in the problem statement.
Your third line should be $P(P_3 \mid F_1 \cap F_2)$, since the third test is attempted only if both the first and second tests are failed.
Your final answer should be
$$P(F_1)\cdot P(F_2\mid F_1)\cdot P(P_3 \mid F_1 \cap F_2) $$
Is that clear?