# Binomial probability distribution

What exactly is binomial distribution?

Q. A manufacturing process is estimated to produce $5\%$ nonconforming items. If a random sample of the five items is chosen, find the probability of getting two nonconforming items.

Now one could say let there be $100$ items. Then the required probability woud be $\frac{{5\choose 2}{95\choose 3}}{{100\choose 5}}$. In what order the items are chosen is irrelevant. This roughly comes out to be $0.18$, while the answer is $0.22$. Where did we go wrong?

Why should we assume there are $100$ items in total? Let us assume $n\to\infty$, as we determine $\frac{{.05n\choose 2}{.95n\choose 3}}{{n\choose 5}}$. What if $0.95 n$ and $0.05n$ are not integers? We use the gamma function.

We get $\frac{{.05n\choose 2}{.95n\choose 3}}{{n\choose 5}}=\frac{\int_{0}^{\infty}{t^{0.05n}e^{-t} dt}.\int_{0}^{\infty}{t^{0.95n}e^{-t} dt}}{{n\choose 5}}$

My textbook says this tends to ${5\choose 2}(0.05)^2 (0.95)^2$. This is something you could verify for yourself.

Another question. Say you roll a die 5 times. Find the probability of getting two $6$s. The probability as determined by combinatorics is $\frac{{5\choose 2}5^3}{6^5}$. You must have applied the binomial theorem before in such problems. You know the answer to be ${5\choose 2}(\frac{1}{6})^2 (\frac{5}{6})^3$. This matches with the answer determined before. So why is it that we’re right here in determining the probability accurately, while we were not before?

Binomial probability corroborates with elementary probability where separate arrangements of selected items are counted as distinct arrangements, and where the total number of items is known and not just guessed at. When the total number of items is not known and only percentages (percentage of success) is known, then binomial probability is an approximation arrived at by assuming $n$ approaches infinity.