I'm experimenting with bayes rule. I have been invited to last stage interview in highly selective company. I want to use bayes rule to evaluate my chances of getting the post.

Let's define the followings :

$J ={}$ The hypothesis that I get the job
$I ={}$ Being in the last stage interview

I want to know $P(J\mid I),$ i.e. the likelihood that I get the job given that I'm in last stage interview. According to Bayes rule: $$ P(J\mid I) = \frac{P(I\mid J) P(J)}{P(I)}$$

My question is how to make sense of $P(I\mid J)$? is it equal to $1$?

Let's say that $P(I)= 0.05,$ on average, from 20 applicants who send their cv, 1 goes to final interview. For the prior $P(J),$ let's say, based on my previous experiences, from 10 jobs I apply to, I get offered 1 position. Then $P(J)$ is $0.1$

then bases on these $P(J\mid I) = 0.1/0.05 = 2$ ok that's a likelihood, it can be greater than $1.$ But how do I interpret it? For me, the problem is with $P(I\mid J)$ but I don't know how to fix it? maybe my formulation is incorrect.
2nd question, for another candidate that has been also invited to the last stage interview, only $P(J)$ changes. is that correct? But $P(I)$ should change as $P(J)$ changes as it can be conditioned on $J.$

Last question, Let's say that I also know from previous experiences that when I get invited to the last stage interview, I get the position 80% of the time. But that's is $P(J\mid I)$ itself. How do I use this information for this company which is more selective (lower $P(I)$). How to incorporate $0.8$ in my prior $P(J)$? obviously I can't say $P(J)=0.8$

  • 1
    $\begingroup$ Hi: The event I is contained in the event J so P(I given J ) = 1. $\endgroup$
    – mlofton
    Commented Dec 15, 2018 at 14:30

2 Answers 2


Maybe I didn't understand the question correctly, but for the first part would try an answer: I think, according to the 'drug testing'-example here: https://en.wikipedia.org/wiki/Bayes_theorem , the demoninator (PI) of the formula should include the chances of the complementary events. Then the formula would be: P(J|I)=0.05*0.1/(0.05*0.1+0.95*0.9)=0.0058.


$P(J|I) $ is a probability, so if it is greater then one, it's a clear sign that you are doing something wrong.

You used inconsistent definitions. It doesn't make much sense if probability of getting the job is higher, then the probability of getting invited to job interview. On another hand, if it would be possible to get the job without the interview, then $P(I|J) < 1$. If in doubt if the numbers make sense, draw a $2 \times 2$ table

$$ \begin{array}{cc} & {\text{not hired}} & {\text{hired}} \\ {\text{not interviewed}} & ? & ? \\ {\text{interviewed}} & ? & ? \\ \end{array} $$

and fill it with counts, e.g. how many out of 100 applicants go to each cell. Each cell by total is joint probability. Cell divided by row (or column) sum is conditional probability. Column (or row) sum by total is marginal probability. This will help you with checking if the numbers add up.

If $P(I)$ is how often applicants get invited to job interviews, $P(J)$ needs to be how often applicants get hired. This is important if you want to talk about conditional probabilities, because probability that someone got to the final interview given that you applied is meaningless.

  • $\begingroup$ Thanks for your answer. my question is about my chances getting hired in the particular company. what would P(I|J) then be? $\endgroup$
    – Lu Yin
    Commented Dec 15, 2018 at 19:32
  • $\begingroup$ @LuYin what kind of data do you have? From what you were saying, you were mentioning probabilities of others being hired in this company, or you in other companies. Neither of those relates directly to your case, so the question which one would you like to use as an approximation. In both cases, you need to be precise, conditional probability of someone getting job in this company conditional on you applying to other companies does not make much sense. $\endgroup$
    – Tim
    Commented Dec 15, 2018 at 19:53

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