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If $X$ is the number of successes in a Binomial Distribution

Then, $P(X \leq K) = \dbinom{n}{k} (1-p)^{(n-k)}$

However, when I apply it to the case where $p=0.003$ $n=1000$ $k=1$

I get $P= 50.43$!!!

What am I missing??

user966168
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  • You don’t have to look that hard for an example that shows that something’s wrong: your formula says that if you toss a fair coin twice, you’re certain to get at most one head! $\binom21\left(1-\frac12\right)^{2-1}=1$. You’re also certain to get at most one tail, so you’re certain to get one of each. – Brian M. Scott Mar 19 '13 at 15:22
  • But my question is how can a probability be > 1? – user966168 Mar 19 '13 at 15:30
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    It can’t: the fact that your calculation produces a result greater than one proves that your calculation is incorrect. Similarly, the fact that it ‘shows’ that tossing two fair coins must result in one head and one tail, something that is obviously false, shows that your calculation is incorrect. – Brian M. Scott Mar 19 '13 at 15:40
  • It is not my calculations, it is taken from a textbook. That is why I was confused. – user966168 Mar 19 '13 at 16:44

1 Answers1

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It should be $P(X = k) = \dbinom{n}{k} p^k(1-p)^{n-k}$ and hence $P(X \leq k) = \sum_{r=0}^k\dbinom{n}{r} p^r(1-p)^{n-r}$.

user1551
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  • That is true, but if you simplify the later you get $P(X \leq K) = \sum_{r=0}^k \dbinom{n}{r}p^r(1-p)^{n-r} \leq \dbinom{n}{n-k}(1-p)^{(n-k)}=\binom{n}{k}(1-p)^{(n-k)}$ – user966168 Mar 19 '13 at 15:22
  • @user966168: So what? Even if it’s correct, which I’ve not checked, that’s an inequality, not an equality. In your example it says that $P(X\le 1)$ is no bigger than about $50$. Since we already know that it’s less than $1$, that information is both correct and rather useless. – Brian M. Scott Mar 19 '13 at 15:42
  • @user966168 Your simplification only shows that $P(X\le k)$ cannot exceed ${n\choose k}(1-p)^{n-k}$. It does not show that $P(X \leq K)$ is equal to ${n\choose k}(1-p)^{n-k}$. – user1551 Mar 19 '13 at 16:00
  • I was saying the same thing. Someone used it in a peer reviewed paper as an approximation and I though there could be an explanation. Anyways, thanks for the comments. – user966168 Mar 19 '13 at 16:46