# Probability of getting a job when applying for 3 places

I have the following problem.

You apply for jobs and know that if you send your application then every job appointment procedure has two stages:

1. You can be either invited or not invited to a personal interview and then
2. You can either be selected or not selected for a job.

Assuming that you do not have any further information about the process and the selection criteria, please compute the a priori probability to get at least one job offer after sending your application to three different places.

### My thoughts

I started thinking about what events are independent and which ones are dependent.

I thought that being able to get a job from place $A$ is an independent event of being able to get a job from place $B$. For this reason, I decided to consider independently the problem what is the probability of getting a job from a place $X$.

Now, the probability of being called to go to the interview (lets call this event $I$) is $\frac{1}{2}$, in other words $P(I) = \frac{1}{2}$, because we can either be called or not.

Then, the probability of getting the job (lets call this event $J$) strictly depends on event $I$, because, for example, you cannot get the job, if you don't first go to the interview.

What we actually want to know is the probability of being called and getting the job, in other words we want to know $P(I \text{ and } J)$. Since these events are dependent, we can use the rule that $P(A \text{ and } B) = P(A) \cdot P(B | A)$, where $P(B|A)$ is the probability of event $B$ happens given the fact that event $A$ has happened.

Applying this rule to my case, I need to find $P(J | I)$, because I already know $P(I) = \frac{1}{2}$. I was thinking that this probability is $\frac{1}{4}$. Why? Basically, from $\frac{1}{2}$ of the possibilities remaining we have half of the chances to get the job, so $\frac{1}{2}$ of $\frac{1}{2}$ is $\frac{1}{4}$.

I can now calculate $P(I \text{ and } J)$, which should be $P(I) \cdot P(J | I) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$.

If my reasonings are correct, this should represent the probability of getting job from one place. Since I am applying for three different jobs (which are not dependent between each other), then I have more possibilities than $\frac{1}{8}$, so I thought we could sum the possibilities of getting a job for each individual place, thus my answer would be $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$.

What am I doing wrong, what am I doing correct? Or can I improve something?

• Like you said, there is not a whole lot of info to get some probability so your set up isn't bad, however one can take issue with that 1/2 about whether or not to be invited. That's like saying if there are 99 black balls and one white ball in the bowl, chances of getting the white is 1/2, because you either draw a white ball or not. In other words, there are lots more of other people sending out their resume... – imranfat Sep 23 '15 at 15:53
• @imranfat Yeah, I hadn't considered that possibility, but in any case what we can assume here is that we have 1/2 of the possibilities to be called or not... – nbro Sep 23 '15 at 15:56
• Everything is ok, but the last sentence with numbers is wrong. What if you applied for 8 jobs, you think that the probability of getting job would be $1$? Or better, what if you applied for the 10 jobs with the same probability? – Mesmerized student Oct 3 '15 at 13:56
• @DanijelSubotic That's exactly what does not make sense for me... We need some how to say that we are applying for 3 jobs... – nbro Oct 3 '15 at 14:29
• I see complementary counting... Find the probability of not getting a job at all first. Then subtract from 1. – user276580 Oct 4 '15 at 22:41

I think I find this line of reasoning rather too speculative, I'm afraid. :-)

First, you assert that $P(I) = 1/2$. I don't see any symmetry that justifies this kind of application of the principle of indifference; it's not as though the only difference between being selected for an interview and not being selected is the opportunity to go to the interview. From the job poster's perspective, they have limited time to interview candidates, whether five people apply for the posting, or five hundred.

But suppose we put that aside for the moment. You then assert that $P(J \mid I) = 1/4$, on the assumption that once half of the possibilities have been eliminated (presumably, the $\neg I$ portion), only half of the half remain, or $1/4$. But $P(J \mid I)$ is already a conditional probability—it expresses the probability of the more specific compound event $I$ and $J$ as a fraction of the probability of the condition $I$! If you are to apply the principle of indifference again, you should treat the two possibilities equally; that is, $P(J \mid I) = P(\neg J \mid I)$. And since, by excluded middle, $P(J \mid I)+P(\neg J \mid I) = 1$ necessarily, it should be the case that

$$P(J \mid I) = P(\neg J \mid I) = \frac{1}{2}$$

$$P(J \mid I) = \frac{1}{4}$$

which implies

$$P(\neg J \mid I) = \frac{3}{4}$$

This has no more basis than the assertion that $P(I) = 1/2$, and violates the principle of indifference to boot. (Of course, I said that the principle of indifference shouldn't be applied here, but since you appear to want to do it...)

In other words, when you arrive at $1/4$, you are determining the probability that you get the interview and you get the job (and that's $P(I, J)$), not the probability that you get the job given that you got the interview (that's $P(J \mid I)$). If you want to analyze the problem that way, there's nothing wrong with that (modulo indifference), but then you should just leave that value alone, and not multiply, again, the probability that you get the interview. That's already been accounted for in the joint probability.

I think the most we can say about it is as follows: Suppose the probability of being selected for a given job posting is $\sigma$. Then, by independence, the probability of being selected for at least one of two job postings is $1-(1-\sigma)^2$; the probability of being selected for at least one of three job postings is $1-(1-\sigma)^3$; and the probability of being selected for at least one of $k$ job postings is $1-(1-\sigma)^k$. Even independence seems rather daring, but I can see a case for it better than I can for indifference.

• First, you should have read better the post: Assuming that you do not have any further information about the process and the selection criteria, please compute... – nbro Oct 9 '15 at 22:18
• @Rinzler: I did read that part, perfectly well, thank you, and I still disagree with applying the principle of indifference. Just because you have no further information doesn't mean there's a basis for picking any particular value. – Brian Tung Oct 9 '15 at 22:20
• And anyway, my answer still gives you a way to proceed from one interview to multiple interviews, should you arrive at an answer for the former that you are satisfied with. – Brian Tung Oct 9 '15 at 22:21
• "Apparently", my reasoning was wrong, the $\frac{1}{4}$ part. Here's the possible options with the respective probabilities. Option 1: you get the interview, but you don't get the job (probability $\frac{1}{4}$). Option 2: you get the interview and the job (probability $\frac{1}{4}$). Option 3: you get the job, but not the interview (probability $0$, because this is not even possible). Option 4: you don't get the interview and the job (probability $\frac{1}{2}$). Summing these probabilities we obtain $1$. – nbro Oct 9 '15 at 22:28
• So, I was wrong in assuming probability $\frac{1}{4}$ for the second part of the selection process. – nbro Oct 9 '15 at 22:28

Assuming that the numbers you use are correct ($1/2$ and $1/4$) everything if pretty fine, except the last sentence and the first sentence of your thoughts.

Let $A$-is the getting job from the first company, $B$ - from the second (for now just with two jobs). Then you just do this: $$P\{A\cup B\}=P\{A\}+P\{B\}$$

And here the problem rises, you can't just split them, because these events have some intersections. So, the right way is: $$P\{A\cup B\}=P\{A\}+P\{B\}-P\{AB\}.$$

And now we come to the first sentence where you assume that these events are independent, but are they? The answer would be YES, but it's not so trivial.

As you already mention $P\{AB\}=P\{A\}P\{B|A\}=\frac{1}{8}P\{B|A\}.$ Now we need to count probability of getting job B, if we now that we get job A? Right here it is more clear (at least for me) that these events are really independent. $$P\{AB\}=\frac{1}{64}=P\{A\}P\{B\}.$$

And now we can calculate for two jobs: $$P\{A\cup B\}=\frac{15}{64}$$

Let's now say that we have $n$ independent jobs to apply at. Assume that for all jobs we have same numbers as you mention above. We have inclusion-exclusion formula (https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle):

$$P\{\bigcup_{k=1}^nA_k\}=\sum_{k=1}^nP\{A_k\}-\sum_{1\leq k<l\leq n}P\{A_kA_l\}+...+(-1)^{n-1}P\{A_1A_2...A_n\}.$$

That's the most general case and in your exact with the given numbers and $n=3$ we can calculate, $C$- event of getting the third job:

$$P\{A\cup B\cup C\}=\frac{3}{8}-{3\choose 2}\frac{1}{64}+\frac{1}{512}=\frac{169}{512}<\frac{3}{8}=\frac{192}{512}.$$

I hope now it's a little bit clearer to you.

• How can you explain that this solution takes into account that we are finding the probability of getting at least one job? – nbro Oct 3 '15 at 14:48
• @Rinzler, it's pretty obvious, but we can show it, mathematically too. What is the event that is opposite to getting at least one job? Of course it's getting now jobs at all. So this event is $\{A\cup B\cup C\}^c=\{A^c \cap B^c\cap C^c\}=\{A^c B^c C^c\}$, where with the "$c$" we denote the complement. Now we know that the events are independent so: $$P\{A^c B^c C^c\}=\frac{7^3}{8^3}=\frac{343}{512}=1-\frac{169}{512}.$$ Which proves that we get the prob of at least one job. – Mesmerized student Oct 3 '15 at 14:55
• I think I got confused between events being independent and mutually exclusive, which are different concepts... – nbro Oct 3 '15 at 15:24

Edit. Your reasoning is almost perfect for the single application case. I would say that $P(J|I) = \frac 12$ if there is equal chance that you do or don't get the job once you are in the interview, thus yielding $P(J) = P(J|I)P(I) + P(J|I^c)P(I^c) = \frac 12 \cdot \frac 12 + 0$ by the law of total probabilities.

Anyway, I will stick with the $1/8$ for this explanation.

What went wrong? The problem with freely adding the $1/8$s was that the $1/8$ chance of getting the first job didn't take into account what happens with the other applications, so it includes by default the scenarios where you get this job and one of the others, this job and two of the others, etc.

However, you can still use what you already calculated. A little more notation helps in making the statements more explicit:

Solution. Let $J$ be the total amount of job offers that you get. Additionally, let $X_1, X_2$ and $X_3$ be random variables that represent the amount of job offers from the first, second and third places, thus taking values in $\{0,1\}$.

With this notation, the probability of getting at least one job is

$$P(J>0) = P(J=1) + P(J=2) + P(J=3).$$

Note that you can add these probabilities up because these events are mutually exclusive. Using your calculations, there is a $1/8$ chance to get a single job, so, assuming that the $\{X_i\}$ are independent: $$P(J=1) = \sum_i P(X_i = 1, X_j=0, X_k=0) = 3\cdot\frac18\cdot\frac78\cdot\frac78 = 3\cdot\frac{7^2}{8^3}.$$

Here $j$ and $k$ are loose notation for "the other two indexes in $\{1,2,3\}$." Similarly,there are three ways of getting a single rejection, so $$P(J=2) = \sum_i P(X_i = 0, X_j=1, X_k=1) = 3\cdot\frac78\cdot\frac18\cdot\frac18 = 3\cdot\frac{7}{8^3},$$ and, as there is only one way to get three offers, $$P(J=3) = \frac18\cdot\frac18\cdot\frac18 = \frac{1}{8^3}.$$

Finally, the result is $$P(J>0) = \frac{3\cdot7^2+3\cdot7+1}{8^3} = \frac{169}{512} .$$

In general. It may help you to know that the variable $J$ which we defined follows a binomial distribution with $p=1/8$ and $n=3.$

If you want to know the answer for $p=1/4$ (or any other probability for getting a single job) or for a different number $n$ of applications you can ask Wolfram Alpha. (Pretty neat, huh?)