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Find the probability of $x$ happening for the first time after $n = 1000$ trials with $p = 0.001$

So after $n$, I'm assuming that this means at most $n$? I know the generic way of solving at most which would be:

$$p(0) + p(1) + \cdots + p(n)$$

but without using $\sigma$ how would I find the sum for a large $n$?

Lee David Chung Lin
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user3538161
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  • Hint: saying that the first hit occurs after $n$ trials just means that the first $n$ trials were all misses. – lulu Oct 24 '15 at 14:24
  • There seems to be an ambiguity in the question -- I would have interpreted it to mean that x happens for the first time exactly after $n$ trials, whereas @lulu seems to have interpreted it to mean "after at least $n$ trials". – joriki Oct 24 '15 at 14:45
  • @joriki That is indeed how I read it, though revisiting the question I think I might agree that your reading is more likely. To be clear, you think the first hit is meant to occur on the $(n+1)^{st}$ trial or on the $n^{th}$? – lulu Oct 24 '15 at 14:53
  • @lulu: I would have thought on the $n$-th, but I agree it's not clear. – joriki Oct 24 '15 at 15:22

1 Answers1

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To summarize the discussion between @joriki and myself in the comments:

The question is somewhat ambiguous. It could mean (at least) one of three things:

I. The first $n$ trials are all misses (no condition on what comes after).

II. The first hit occurs on the $n^{th}$ trial (so that the first $n-1$ trials are all misses and the $n^{th}$ is a hit).

III. The first hit occurs on the $(n+1)^{st}$ trial (so that the first $n$ trials are all misses and the $(n+1)^{st}$ is a hit).

No way to resolve the ambiguity (other than going back to the original source for clarification). But the solutions are fairly similar. Let $p$ be the probability of a hit and $q=1-p$ be the probability of a miss. (in the OP $p=.001$ but we can work in general with no extra effort).

The probability of two independent events both occurring is the product of the individual probabilities. Hence:

I. the probability is $q^n$.

II. the probability is $q^{n-1}p$.

III. the probability is $q^{n}p$.

lulu
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