The Binomial Distribution

When is a setting binomial? (BINS)

A situation is binomial when all four hold:

• Binary: each trial is a success or a failure.
• Independent: trials don't affect each other.
• Number of trials $n$ is fixed in advance.
• Same probability $p$ of success on every trial.

Then $X$ = the number of successes is a binomial random variable.

The binomial formula

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

Reading it: $\binom{n}{k}$ counts which $k$ of the $n$ trials are the successes (that's the combinations from counting), $p^k$ is the probability those $k$ succeed, and $(1-p)^{n-k}$ is the probability the rest fail.

Worked example

Flip a fair coin 5 times ($n = 5$, $p = 0.5$). Probability of exactly 3 heads:

$P(X = 3) = \binom{5}{3}(0.5)^3(0.5)^2 = 10 \cdot 0.125 \cdot 0.25 = 0.3125$

With $p \neq 0.5$. A shooter makes free throws $80\%$ of the time ($p = 0.8$). In 6 attempts, probability of exactly 5 makes:

$\binom{6}{5}(0.8)^5(0.2)^1 = 6 \cdot 0.32768 \cdot 0.2 \approx 0.393$

"At least" problems — use the complement

For "at least one," it's almost always easier to subtract from 1.

Example. Roll a die 4 times. Probability of at least one 6:

$P(\text{at least one } 6) = 1 - P(\text{no } 6) = 1 - \left(\tfrac{5}{6}\right)^4 \approx 1 - 0.482 = 0.518$

Mean and standard deviation

A binomial distribution has a tidy mean and spread:

$\mu = np \qquad \sigma = \sqrt{np(1 - p)}$

Example. $n = 100$ fair flips ($p = 0.5$): $\mu = 50$ and $\sigma = \sqrt{100 \cdot 0.5 \cdot 0.5} = \sqrt{25} = 5$.

Normal approximation

When $n$ is large — a common rule of thumb is $np \geq 10$ and $n(1-p) \geq 10$ — the binomial is approximately normal with that same $\mu$ and $\sigma$. So the 68-95-99.7 rule kicks in. For 100 fair flips, about $95\%$ of the time you'd get between $40$ and $60$ heads ($50 \pm 2 \cdot 5$).

Key takeaways

• Binomial setting = BINS: binary, independent, fixed $n$, same $p$.
• $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.
• For "at least one," compute $1 - P(\text{none})$.
• $\mu = np$ and $\sigma = \sqrt{np(1-p)}$.
• Large $n$ ($np$ and $n(1-p) \geq 10$) makes it approximately normal.