I'm trying to calculate the probability of exactly $8$ identical independent events occurring if each has a $25\%$ chance of occurring.
To simplify, if I were rolling a four-sided die $8$ times, what would be the probability be of rolling a $1$ only once through the probability of rolling a $1$ on the die $8$ times? I'd like to create a data table with $x$ being $1$ through $8$ and $f(x)$ being the probability. It seems like a normal distribution.
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$\begingroup$It follows a binomial distribution.
If $X$ is distributed according to the binomial distribution with number of trials $n$ ($8$, in your case) and probability $p$ ($0.25$ for you), then $X\sim\mathcal {Bin}(n,p)$ means:
$$\mathsf P(X=r) \,=\, \binom{n}{r}\,p^r\,(1-p)^{n-r} \qquad \text{where }r\in\{0,..,n\}, 0\text{ otherwise}$$
Where $\dbinom{n}{r} = \dfrac{n}{r!\,(n-r)!}$, which is also often represented as: ${^n{\rm C}_r}$
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