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The probability that a device passes the quality control test is $0.8$. Find the probability that a given device will pass the test on the third try.

I don't understand how to go about answering this question. It seems to me like there isn't enough information. Do I assume each trial is independent? If so, I thought to use the probability distribution of a geometric RV, but this PMF gives the probability of obtaining the first success at the kth Bernoulli trial. The question says nothing about it having to be the first success. So is the answer just $0.8$ since the probability is the same for any trial?

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Well, I suppose you're right, there isn't enough information given in the question. This isn't a problem because we could help fill in what information could have been contained in the question, so we could improvise and see what different scenarios there are. (Of course, this question doesn't require any PMFs or distributions to help us find the answer.)

If we suppose we're just retesting this device regardless of whether it passed the previous tests or not, then the probability of the device passing this third test is simply $0.8$.

If we suppose that we're retesting the device because the previous tests failed, then the probability of not passing the first, $0.2$, multiplied by the probability of not passing the second test, $0.2$, (supposing these events are independent), multiplied by the probability of passing the third test, $0.8$, gives the required probability: $0.032$.

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  • $\begingroup$ 0.032 seems to be the consensus. I noticed that the pmf of the geometric Rv also gives 0.032 with p=0.8 and k=3 which suggests that this is indeed the correct interpretetation of the question given that it is under a Random variable question bank and the other interpretation would be too trivial. $\endgroup$
    – user492613
    Oct 17, 2017 at 18:31
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Simply 0.2 * 0.2 * 0.8 = 0.032 is the probability of the first AND second tests will be failed AND the third will be passed.

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