2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ They aren't disjoint, if you fail the first, there is no second interview. $\endgroup$
    – Macavity
    Dec 10 '17 at 21:10
7
$\begingroup$

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:

$$P(F_2|F) = \frac{P(F_2 \cap F)}{P(F)} = \frac{0.18}{0.496}$$

$\endgroup$
1
$\begingroup$

There are three ways to fail, with probabilities $.1$, $(.9)(.2)=.18$, and $(.9)(.8)(.3)=.216$. That gives a total probability $.496$ of failing, of which $.18$ comes from the event we’re interested in. Thus:

$$\frac{.18}{.496}=\frac{45}{124}\approx .3629$$

The individual probabilities are more complicated because you can only fail the second interview if you pass the first one, etc.

$\endgroup$
0
$\begingroup$

Q: "If she didn’t get the job, what is the probability that she failed the second interview?"

Zero, or 0.8 if three people interview her simultaneously.

  • "She passes the first, second, and third interviews ...".

  • "If she fails any interview, she cannot continue with subsequent interview(s) and will not get the job.".

Since she passed the second interview, as evidenced by the third interview (and your assertion that she passed), there's zero change she failed the second interview.

If three people sit in on the interview and take turns asking questions, afterwards the interviewer designated as first tells the second person that they can tally their results, then the probability is as you stated it 0.8.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.