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Assuming every trial is independent from all the others and the probability of a successful run is the same every trial, how can you determine the chance of a successful trial a set number of times or more?

For example, You run 20 independent trials and the chance of a "successful" independent trial each time is 60%. how would you determine the chance of 3 or more"successful" trials?

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Do these three successful trials have to happen in a row? – Edmon Aug 4 '12 at 17:07
No. I'm sorry I wasn't more specific. The successful trials do not have to occur in any particular order. – Franks Aug 4 '12 at 17:25
up vote 4 down vote accepted

If the probability of success on any trial is $p$, then the probability of exactly $k$ successes in $n$ trials is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ For details, look for the Binomial Distribution on Wikipedia.

So to calculate the probability of $3$ or more successes in your example, let $p=0.60$ and $n=20$. Then calculate the probabilities that $k=3$, $k=4$, and so on up to $k=20$ using the above formula, and add up.

A lot of work! It is much easier in this case to find the probability of $2$ or fewer successes by using the above formula, and subtracting the answer from $1$. So, with $p=0.60$, the probability of $3$ or more successes is $$1-\left(\binom{20}{0}p^0(1-p)^{20}+\binom{20}{1}p(1-p)^{19}+\binom{20}{2}p^2(1-p)^{18} \right).$$

For the calculations, note that $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. In particular, $\binom{20}{0}=1$, $\binom{20}{1}=20$ and $\binom{20}{2}=\frac{(20)(19)}{2!}=190$.

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