The binomial distribution: count the paths, multiply the chances

Binomial distribution

Where does P(X = k) = C(n, k) pᵏ(1 − p)ⁿ⁻ᵏ come from, and why is the mean np?

Repeat a trial n times with success chance p. Each sequence with k successes has the same probability pᵏ(1 − p)ⁿ⁻ᵏ, and there are C(n, k) of them. The board adds up the bars to measure the total, the mean and the variance, and compares them with 1, np and np(1 − p).

High school
Independent trials
Binomial probability
Mean np
Variance np(1 − p)

n

p

0.30

k

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

$P(X = k)$: 0.2668
$\binom{n}{k}$: 120
$p^k(1-p)^{n-k}$: 0.002224
$\textstyle\sum P$: 1.000000
$\text{mean of bars}$: 3.0000
$np$: 3.0000
$\text{variance of bars}$: 2.1000
$np(1-p)$: 2.1000

Guided tour

1One path, then all of them
Ten trials, success chance $0.3$ . One particular sequence with exactly $3$ successes, say SSSFFFFFFF, has probability $0.3^3\times0.7^7=0.002224$ because independent chances multiply. There are $\binom{10}{3}=120$ such sequences, all equally likely, so $P(X=3)=120\times0.002224=0.2668$ .
2The bars add to 1
Every sequence has some number of successes between $0$ and $n$ , so the bars share out all the probability: they add to $1$ . The mean of the bars, $\sum k\,P(X=k)$ , comes out $3=np$ , and their variance $2.1=np(1-p)$ .
3A fair coin is symmetric
At $p=0.5$ swapping success and failure changes nothing, so the bars are symmetric about $np=10$ . The tallest is $P(X=10)=0.1762$ for $20$ tosses: even the most likely count happens less than one time in five.
4Rare events
With $p=0.05$ and $20$ trials, $P(X=0)=0.95^{20}=0.3585$ : about a one-in-three chance of no success at all, and the mean is only $np=1$ .

Explanation

1
Where the formula comes from
Write the $n$ results as a string of S and F. Because the trials are independent, the probability of one string is the product of its letters' chances; a string with $k$ S's and $n-k$ F's has probability $p^k(1-p)^{n-k}$ , whatever the order. The number of such strings is the number of ways to choose which $k$ positions hold S: $\binom nk$ . Strings are disjoint outcomes, so their probabilities add:
$P(X=k)=\binom nk\,p^k(1-p)^{n-k}$
2
Why the total is 1
Expand $(p+(1-p))^n$ by multiplying out the $n$ brackets: picking $p$ from $k$ of them and $1-p$ from the rest can be done in $\binom nk$ ways. So the bars are exactly the terms of the binomial expansion, which is where the name comes from, and they add to $1^n=1$ .
3
Mean and variance without the formula
Let $X_i=1$ if trial $i$ succeeds and $0$ if not, so $X=X_1+\cdots+X_n$ . Each $X_i$ has mean $1\cdot p+0\cdot(1-p)=p$ and variance $E(X_i^2)-p^2=p-p^2=p(1-p)$ . Means always add, and for independent trials variances add too:
$E(X)=np,\qquad D(X)=np(1-p)$
The board does not use these: it sums the bars directly, and the results agree.
4
When it applies
Three conditions: a fixed number $n$ of trials, each with two outcomes and the same chance $p$ , and independence between them. Drawing without replacement breaks the last two, since each draw changes what is left; that case is the hypergeometric distribution.

Check yourself

Q01

A fair coin is tossed $4$ times. What is $P(\text{exactly } 2 \text{ heads})$ ?

Q02

Why is every sequence with $k$ successes equally likely?

Q03

A test has $50$ multiple-choice questions with $4$ options each. Guessing every one, what is the expected score?

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