# Binomial Cumulative Distribution probability

If you were to do a thing $x$ times, and 35% of the time it worked, what's the chance that after $x$ times it would have worked 240 or more times?

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That depends on how many trials there were... if at least $240/0.35$, 1; if less, 0. – vonbrand Apr 7 '13 at 17:30
Between 240 and 685. And I'm confused by your comment, 686 times would not lead to a 100% chance, right? – Cool12309 Apr 7 '13 at 17:35
It isn't clear from your question if you are asking about the trials done or a new set of trials... – vonbrand Apr 7 '13 at 17:38
I just edited it. How about now? – Cool12309 Apr 7 '13 at 17:41

Note that you are describing a random variable $X\sim B(x,0.35)$. Using the known PMF (probability mass function) of the binomial distribution we have:

$$P(X=\alpha)={x \choose \alpha}\left(0.35\right)^{\alpha}(0.65)^{x-\alpha}$$

Therefore, we have:

$$P(X\geq 240)=1-P(X\lt 239)=1-\sum_{\alpha=0}^{239}{{x \choose \alpha}\left(0.35\right)^{\alpha}(0.65)^{x-\alpha}}$$

Which depends on the value of your variable $x$.

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Do have a calculator where I could put this in? – Cool12309 Apr 7 '13 at 17:57
You can use this to compute the summation, and then just subtract the answer from $1$ to give you your final result. – Shaktal Apr 7 '13 at 17:58
I tried this but it doesn't seem to be the right formula – Cool12309 Apr 7 '13 at 18:05
@Cool12309 That is because you forgot to include the binomial co-efficient: $$\binom{400}{\alpha}$$ The correct sum is here. – Shaktal Apr 7 '13 at 18:11
I subtracted that from 1, but that would give a crazy low chance. I thought it would be much higher than that? – Cool12309 Apr 7 '13 at 18:15