Binomial Coefficient vs Binomial Distribution

The binomial coefficient $\binom{n}{k}$ (read "n choose k") counts how many subsets of size $k$ can be drawn from $n$ distinct items, which is the same as counting how many arrangements of $k$ successes and $n - k$ failures fit into a sequence of $n$ trials. The formula is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, with $0!$ defined to equal 1 and $k$ a whole number from 0 to $n$.

Take 5 free throws and ask in how many ways exactly 2 of them could be the makes: $\binom{5}{2} = \frac{5!}{2!\,3!} = \frac{120}{2(6)} = 10$. Those 10 orders are what the coefficient contributes to a probability. If the shooter makes 70 percent of her attempts, the chance of exactly 2 makes is $10(0.7)^2(0.3)^3 = 10(0.49)(0.027) = 0.1323$.

The error to name is treating the coefficient as a probability. "$\binom{10}{5} = 252$, so the probability is 252" is wrong twice over: 252 is a count of orders, and no probability can exceed 1. Nothing about chance enters until that count is multiplied by $p^k(1-p)^{n-k}$, which is the probability of one particular order.

The second slip is using the ordered count instead. Picking 2 of 5 shots in order gives $5 \times 4 = 20$ possibilities, but that counts "first and third" separately from "third and first" even though the pair of makes is identical. Dividing by the $2! = 2$ orderings of the same pair returns 10. The $k!$ in the denominator is exactly that correction, and it is the whole difference between a permutation count and a combination count.

Two facts fall straight out of the formula. It is symmetric, $\binom{n}{k} = \binom{n}{n-k}$, because choosing which $k$ trials succeed is the same act as choosing which $n - k$ fail, so $\binom{5}{2} = \binom{5}{3} = 10$. And $\binom{n}{0} = \binom{n}{n} = 1$, since there is one way for every trial to fail and one way for every trial to succeed. The distribution built on it is topic 2.10 of Unit 2.

A binomial distribution counts successes across a fixed number of trials that are binary, independent, and identical in success probability. Write it $X \sim B(n, p)$: $n$ trials, success probability $p$ on each one, and $X$ the number of successes, which can be any whole number from 0 to $n$. The probability of exactly $k$ successes is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $\binom{n}{k}$ (n choose k) counts the orders in which those $k$ successes could fall. The mean is $\mu_X = np$ and the standard deviation is $\sigma_X = \sqrt{np(1-p)}$.

Flip a fair coin 10 times, so $n = 10$ and $p = 0.5$. Then $P(X = 6) = \binom{10}{6}(0.5)^6(0.5)^4 = \frac{210}{1024} = 0.2051$. The mean is $\mu_X = 10(0.5) = 5$ heads and the standard deviation is $\sigma_X = \sqrt{10(0.5)(0.5)} = \sqrt{2.5} \approx 1.581$ heads.

Now the error that costs the most marks: "so the probability of 6 or more heads is 0.2051." It is not. That number is the probability of exactly 6. At least 6 means 6, 7, 8, 9 or 10, and adding those five probabilities gives 0.3770, nearly double. Circle the words at least, at most, more than and fewer than before any arithmetic starts, because the formula answers only the exactly question and every other question is assembled out of it.

Independence is the condition that breaks most often in practice. Deal 10 cards from a deck without replacement and the chance of a heart shifts with every card removed, so the count of hearts is not binomial. It is close enough to treat as binomial when the sample is a small fraction of the population, which is what the 10 percent condition checks.

The binomial distribution is topic 2.10 of Unit 2, Probability, Random Variables, and Probability Distributions.