Binomial distribution

The binomial distribution models the number of successes in $n$ independent trials, each with the same success probability $p$. Examples: number of heads in $n$ flips of a possibly biased coin; number of defective items in a batch of $n$; number of correct guesses in an MCQ test.

Setup

The binomial experiment:

1. There are $n$ trials.
2. Each trial has two outcomes: success (with probability $p$) or failure (with probability $1 - p$).
3. The trials are independent.
4. $p$ is the same for every trial.

Let $X$ = total number of successes in $n$ trials. Then $X$ has a binomial distribution with parameters $n$ and $p$, written $X \sim B(n, p)$.

The probability mass function

For $k \in \{0, 1, 2, \ldots, n\}$: $P(X = k) = \binom{n}{k}\,p^k\,(1 - p)^{n - k}.$

Why? There are $\binom{n}{k}$ ways to choose which $k$ trials succeed. Each specific arrangement of $k$ successes and $n - k$ failures has probability $p^k (1-p)^{n-k}$, since trials are independent.

Sanity check: the values must sum to $1$. $\sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} = (p + (1 - p))^n = 1$ by the binomial theorem.

Mean and variance

For $X \sim B(n, p)$: $E[X] = np, \qquad \mathrm{Var}(X) = np(1 - p).$

Why? Write $X = X_1 + X_2 + \cdots + X_n$ where $X_i$ indicates whether the $i$-th trial succeeded. Each $X_i$ is a Bernoulli($p$) variable: $P(X_i = 1) = p$, $P(X_i = 0) = 1 - p$, so $E[X_i] = p$, $\mathrm{Var}(X_i) = p(1 - p)$. By linearity of expectation: $E[X] = np$. By independence of trials: $\mathrm{Var}(X) = np(1 - p)$.

Common probabilities

• Exactly $k$ successes: $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$.
• At least $k$ successes: $P(X \ge k) = \sum_{j=k}^n \binom{n}{j}p^j(1-p)^{n-j}$.
• At most $k$ successes: $P(X \le k) = \sum_{j=0}^k \binom{n}{j}p^j(1-p)^{n-j}$.
• None succeed: $P(X = 0) = (1 - p)^n$.
• At least one succeeds: $P(X \ge 1) = 1 - (1 - p)^n$.

Worked examples

Example 1. A coin is biased: $P(H) = 0.6$. Flipped $5$ times. $P(\text{exactly 3 heads})$?

$P(X = 3) = \binom{5}{3}(0.6)^3(0.4)^2 = 10 \cdot 0.216 \cdot 0.16 = 0.3456$.

Example 2. A multiple-choice quiz has $10$ questions, each with $4$ options. A student guesses every answer. $P(\text{at least 4 correct})$?

$X \sim B(10, 0.25)$. $P(X \ge 4) = 1 - P(X \le 3) = 1 - \sum_{k=0}^{3}\binom{10}{k}(0.25)^k(0.75)^{10-k}$.

Each term: $P(X = 0) = (0.75)^{10} \approx 0.0563$. $P(X = 1) = 10 \cdot 0.25 \cdot (0.75)^9 \approx 0.1877$. $P(X = 2) = 45 \cdot 0.0625 \cdot (0.75)^8 \approx 0.2816$. $P(X = 3) = 120 \cdot 0.0156 \cdot (0.75)^7 \approx 0.2503$.

Sum $\approx 0.7759$. So $P(X \ge 4) \approx 0.2241$.

Example 3. $X \sim B(8, 0.5)$. Find $E[X]$ and $\mathrm{Var}(X)$.

$E[X] = 8 \cdot 0.5 = 4$. $\mathrm{Var}(X) = 8 \cdot 0.5 \cdot 0.5 = 2$.

Example 4. A factory produces items, $5\%$ defective. A box has $20$ items. $P$(at most $2$ defective)?

$X \sim B(20, 0.05)$. $P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$.

$P(X = 0) = (0.95)^{20} \approx 0.3585$. $P(X = 1) = 20 \cdot 0.05 \cdot (0.95)^{19} \approx 0.3774$. $P(X = 2) = 190 \cdot 0.0025 \cdot (0.95)^{18} \approx 0.1887$.

Sum $\approx 0.9246$.

Example 5. A drug works in $80\%$ of patients. Treated $5$ patients. $P$(all 5 recover)?

$P(X = 5) = (0.8)^5 = 0.3277$.

Example 6. Find $n$ so that for $X \sim B(n, 0.3)$, $P(X \ge 1) \ge 0.95$.

$P(X \ge 1) = 1 - (0.7)^n$. Want $1 - (0.7)^n \ge 0.95$, so $(0.7)^n \le 0.05$. Take logs: $n \ln 0.7 \le \ln 0.05$, so $n \ge \ln 0.05 / \ln 0.7 = -2.996 / -0.357 \approx 8.4$. So $n = 9$.

Try it yourself

1. $X \sim B(6, 0.5)$. Find $P(X = 2)$.
2. $X \sim B(10, 0.3)$. Find $E[X], \mathrm{Var}(X)$.
3. Flip a fair coin $4$ times. $P(\text{exactly 2 heads})$.
4. $X \sim B(5, 0.4)$. Find $P(X \le 1)$.
5. A biased die ($P(\text{6}) = 1/6$) is rolled $4$ times. $P(\text{at least one 6})$.
6. $20\%$ of items are defective. Sample of $10$. $P(\text{exactly 2 defective})$.
7. MCQ with $20$ questions, $4$ options. Random guessing. $P(\text{at least 5 correct})$.
8. A basketball player has $80\%$ free-throw rate. Shoots $5$ times. $P(\text{exactly 4 successful})$.
9. A drug cures $70\%$ of patients. Among $10$, $P(\text{all 10 cured})$.
10. $X \sim B(n, p)$ with mean $4$ and variance $3$. Find $n$ and $p$.
11. A coin is flipped $7$ times. Find probability of "at least one heads and at least one tails."
12. A factory: $1\%$ defective rate. Box of $50$ items. Probability of no defective items.
13. Probability of rain on any given day is $0.4$. Over a week (7 days), probability of rain on at least 5 days?
14. A salesman makes $4$ sales calls per day, each with $30\%$ chance of closing. $P(\text{exactly 2 sales})$ on a given day.

Pitfalls and tricks

• Identify the experiment as binomial first. Trials independent? Two outcomes? Same probability throughout?
• $\binom{n}{k}$ is the binomial coefficient, often called "$n$ choose $k$": $\binom{n}{k} = \dfrac{n!}{k!(n - k)!}$.
• Computing $P(X \ge k)$ via complement is often faster: $1 - P(X \le k - 1)$.
• $E[X] = np$ and $\mathrm{Var}(X) = np(1-p)$ , memorise these. Note $E[X] > \mathrm{Var}(X)$ always (since $1 - p < 1$).
• Watch for boundary cases like "at least one" = $1 - (1-p)^n$. Quick mental shortcut.