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I have posted a link there to this topic so the OP can see my work.A die is thrown 60 times. What is the probability of 10 or fewer "sixes" ?

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I have posted a link there to this topic so the OP can see my work.A die is thrown 60 times. What is the probability of 10 or fewer "sixes" ?

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We want to use the binomial probability formula:

\(\displaystyle P(x)={n \choose x}p^x(1-p)^{n-x}\)

where:

\(\displaystyle 0\le x\le10\in\mathbb{Z}\)

\(\displaystyle n=60\) is the number of trials

\(\displaystyle p=\frac{1}{6}\) is the probability of rolling a six in any trial.

We want to find the probability that we get zero 6's or 1 6 or 2 6's or...10 6's. So we want to sum up the probabilities for $x=0$ to $x=10$. With the aid of technology, we find:

\(\displaystyle P(X)=\sum_{x=0}^{10}\left({60 \choose x}\left(\frac{1}{6} \right)^x\left(\frac{5}{6} \right)^{60-x} \right)=\frac{9504082209854658458425546996295452117919921875}{16291225993563085829774250757924867955220283392}\)

\(\displaystyle P(X)\approx0.583386555045634235343489438735516181052152146974030836906911\)

I entered the command:

sum of nCr(60,k)(1/6)^k(5/6)^(60-k) for k=0 to 10

at Wolfram|Alpha: Computational Knowledge Engine