Binomial distribution

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.