The question is as followed:
"A college graduate is applying for a job and has 3 interviews with Google. She passes the first, second, and third interviews with probability 0.9, 0.8, and 0.7, respectively. If she fails any interview, she cannot continue with subsequent interview(s) and will not get the job. If she didn’t get the job, what is the probability that she failed the second interview?"
For me this a problem of finding out probability $P(F_2|F)$, or in other words: the probability she failed the second interview $(F_2)$ given she didn't land the job $(F)$. So going back to the formula:
$$
P(F_2|F) = P(F_2 \cap F) / P(F)
$$
So, now I was looking for the numbers to plug in. Given that I assumed that all events are disjointed, then the probability of failure to land a job ($P(F)$) would be the sum of the probabilities of failing each independent interview (of which the success possibilities are given). So:
$$
P(F) = .6
$$
And the intersection of $F_2$ and $F$ would be $.2$. So after plugging in the answer I would get is $.33$ but it is wrong. I'm not sure how to solve it.
Best Answer
The probability to fail at all is $1$ minus the probability of passing al interviews, so:
$$P(F) = 1-0.9\cdot 0.8\cdot 0.7 = 1-0.504=0.496$$
The intersection of $F_2$ and $F$ is of course just $F_2$, since if you fail the second interview, then you fail, period. Also, to fail the second interview means that you did pass the first, and hence:
$$P(F_2 \cap F) = P(F_2) = 0.9 \cdot 0.2 = 0.18$$
And so:
$$P(F_2|F) = \frac{P(F_2 \cap F)}{P(F)} = \frac{0.18}{0.496}$$