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Suppose that when you submit a job application you have a probability of $0.1$ to receive an interview, and a job interview results in a job offer with probability $0.2$. Also assume you only submit one application at a time (i.e. you wait to know if you have been rejected from the job, either at the application stage or the interview stage before applying to another job).

What is the probability of getting your first job offer after submitting at most 3 job applications?

Attempt: Either you get a job offer after your first, second, or third application.

  • The probability of getting an offer after your 1st application is $(0.1)(0.2)=0.02$.
  • The probability of getting an offer after your 2nd application is $(1-0.02)(0.1)(0.2)=0.0196$
  • The probability of getting an offer after your 3rd application is $(1-0.02)(1-0.02)(0.1)(0.2)=0.0192$

Thus, $0.02+0.0196+0.0192=0.0588$

I don't think my approach is correct, it doesn't make sense to me that you have a higher probability of getting an offer on your first application than you do on your 2nd or 3rd. I'm stumped as to what the mistake is that I am making.

StubbornAtom
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diteni1927
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    "It doesn't make sense to me that you have a higher probability of getting an offer on your first application than you do on your 2nd or 3rd" The numbers you calculated are not that you get an offer on your second or third but are rather the probabilities that not only did you get an offer on your second or third respectively but also that you failed to get an offer on all prior attempts. – JMoravitz Nov 11 '20 at 23:47
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    As for an easier approach... you got your first offer within the first three attempts iff it is not the case that you did not get an offer in any of the first three attempts. $1 - 0.98^3$ – JMoravitz Nov 11 '20 at 23:50

2 Answers2

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Let $a = 0.1$ the probability to get an interview and $b = 0.2 = 2a$ the probability to be hired after an interview.

The probability to be hired at a given submission is $ab$.

The probability you are looking for is:

$$p = \underbrace{ab}_{\text{1st sub. hired}} + \underbrace{(1-ab)}_{\text{1st sub. fail}}\cdot\left(\underbrace{ab}_{\text{2nd sub. hired}} + \underbrace{(1-ab)}_{\text{2nd sub. fail}}\cdot\underbrace{ab}_{\text{3rd sub. hired}}\right) =\\ = a^3b^3 - 3a^2b^2 + 3ab = \\ = a^3 (2a)^3 - 3a^2 (2a)^2 + 3a(2a) = \\ = 8 a^6 - 12 a^4 + 6 a^2 = \\ = 8 \cdot 10^{-6} - 12 \cdot 10^{-4} + 6 \cdot 10^{2} = 0.058808.$$

the_candyman
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The probability you labeled as that of getting an offer after your 2nd application is the probability that you didn't get an offer after your first application and you got an offer after your second...so your actual probability to get a job after the second application is 0.02+0.0192....

nosuchthingasmagic
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