The binomial distribution

Statistics · Senior school statistics · about 10 min

A binomial model counts how many successes occur in a fixed number of repeated trials. It is useful for questions involving correct answers, passed inspections or successful attempts, provided each attempt has the same success probability and trials are independent.

The key idea

If $X\sim\operatorname{Bin}(n,p)$ , then

P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}

Here $n$ is the number of independent trials, $p$ the probability of success per trial, and $k$ the number of successes.

Draw a discrete bar plot for six independent trials with success probability 0.5. The horizontal axis runs from integer k=0 to k=6, with exactly seven equal-width separated vertical bars. Set the bar-height ratios at k=0 through 6 to 1:6:15:20:15:6:1, so the distribution is symmetric and tallest at k=3. The vertical axis is probability and starts at zero. Label every integer k on the horizontal axis, but do not put numerical probabilities on the bars. — A binomial distribution assigns probability to each possible success count.

Check the model first

Use a binomial model when there is a fixed number of trials, exactly two outcomes per trial, independent trials and a constant success probability $p$ . The variable $X$ counts successes, so it can take the values $0,1,\ldots,n$ . When sampling without replacement from a small population, independence and constant probability may fail, so a binomial model is not exact. The combinations term $\binom{n}{k}=n!/[k!(n-k)!]$ counts the different arrangements of $k$ successes among $n$ trials.

Exact and cumulative probabilities

The formula returns the probability of exactly $k$ successes. For a range, add the relevant exact probabilities, for example $P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)$ . For at least one, the complement is shorter: $P(X\geq1)=1-P(X=0)=1-(1-p)^n$ . Watch inclusive wording: at most $k$ includes $k$ , whereas fewer than $k$ does not. A probability is a number between zero and one; convert to a percentage only when requested.

Mean and spread

The expected number of successes is $E[X]=np$ and the variance is $\operatorname{Var}(X)=np(1-p)$ . Standard deviation is $\sqrt{np(1-p)}$ . The mean need not be an integer because it is a long-run average rather than a required outcome. An increasing $n$ generally increases the expected number of successes at fixed $p$ . A probability histogram has a bar at every integer from zero to $n$ , and all bar heights sum to one. This makes a useful check when calculations seem unreasonable.

Worked example: Successful inspections

Each of $8$ independently inspected components has a $25\%$ probability of passing a demanding test. Find the probability that exactly $2$ components pass. Also determine the expected number of passes.

Draw a discrete probability bar chart for n=8 and p=0.25, with nine bars at k=0,1,2,3,4,5,6,7,8. Make each bar proportional to the exact binomial mass C(8,k)(0.25)^k(0.75)^(8-k). Emphasise only the bar at k=2 using a contrasting fill and a small pointing arrow. The vertical scale is linear and begins at zero. Do not label any probability heights, calculate a probability, or show a final answer. — Exactly two passes corresponds to one bar of the eight-trial distribution.

Show the solution

Let $X$ be the number passing. Independence and constant success probability justify $X\sim\operatorname{Bin}(8,0.25)$ .

Select $n=8$ , $k=2$ , $p=0.25$ and $1-p=0.75$ .
Calculate the number of possible success arrangements:

\binom{8}{2}=\frac{8!}{2!6!}=28

Substitute into the probability formula:

P(X=2)=28(0.25)^2(0.75)^6\approx0.31146

Compute the expected count:

E[X]=np=8(0.25)=2

Answer: $P(X=2)\approx31.15\%$ and the expected number of passes is $2$ .

Common mistakes

Treating unequal trials as binomial

Check both independence and a constant success probability before using the formula.

Omitting the combinations term

Include $\binom{n}{k}$ when computing the probability of exactly $k$ successes.

Confusing at least with exactly

For $P(X\geq1)$ , use $1-P(X=0)$ rather than $P(X=1)$ .

Using variance as standard deviation

Take the square root of $np(1-p)$ to obtain standard deviation.

Check your understanding

1. If $X\sim\operatorname{Bin}(4,0.5)$ , what is $P(X=3)$ ?

2. For $X\sim\operatorname{Bin}(12,0.2)$ , find $E[X]$ .

3. Five independent trials each have success probability $0.10$ . What is the probability of at least one success?

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