The Binomial Distribution

The binomial distribution counts successes in a fixed number of independent trials. Recognising when it applies — and when it does not — matters more than the arithmetic.

When a binomial applies

Four conditions, all required:

1. A fixed number of trials, $n$, decided in advance.
2. Each trial has exactly two outcomes — success or failure.
3. The probability of success, $p$, is constant across trials.
4. The trials are independent.

Then $X$, the number of successes, follows

$$X \sim B(n, p)$$

Qualifies: 20 free throws at a fixed success rate; 15 components tested from a large production run; 10 tosses of a coin.

Does not qualify: drawing 5 cards from a deck without replacement — $p$ changes each draw; counting how many attempts until the first success — $n$ is not fixed; recording exam grades A–F — more than two outcomes.

The probability formula

$$P(X = r) = \binom{n}{r}p^r q^{n-r} \qquad \text{where } q = 1-p$$

The three parts: $\binom{n}{r}$ counts the orders in which $r$ successes can occur, $p^r$ is the probability of those successes, and $q^{n-r}$ the probability of the remaining failures.

For $X \sim B(10, 0.3)$, find $P(X = 3)$.

$$P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.0823543 = 0.267 \text{ (3 s.f.)}$$

For $X \sim B(5, 0.5)$, find $P(X = 2)$.

$$\binom52 (0.5)^2(0.5)^3 = 10 \times (0.5)^5 = \frac{10}{32} = 0.3125$$

Cumulative probabilities

"At most", "at least" and "fewer than" require several terms added — or, far more often, a complement.

For $X \sim B(10, 0.3)$, find $P(X \le 1)$.

$$P(X=0) = (0.7)^{10} = 0.0282$$ $$P(X=1) = \binom{10}{1}(0.3)(0.7)^9 = 10 \times 0.3 \times 0.040354 = 0.1211$$ $$P(X \le 1) = 0.0282 + 0.1211 = 0.149 \text{ (3 s.f.)}$$

And $P(X \ge 1)$?

$$P(X \ge 1) = 1 - P(X = 0) = 1 - 0.0282 = 0.972$$

— one term instead of ten.

Mean and variance

$$E(X) = np \qquad \operatorname{Var}(X) = npq$$

For $X \sim B(10, 0.3)$:

$$E(X) = 10 \times 0.3 = 3 \qquad \operatorname{Var}(X) = 10 \times 0.3 \times 0.7 = 2.1$$

The standard deviation is $\sqrt{2.1} = 1.45$.

The mean is intuitive: 10 trials at a 30% success rate should give about 3 successes.

The shape of the distribution

• $p = 0.5$ — symmetric about $\frac{n}{2}$.
• $p < 0.5$ — positively skewed, bunched towards the low values.
• $p > 0.5$ — negatively skewed, bunched towards the high values.

As $n$ grows the distribution becomes more symmetric whatever $p$ is, which is why a binomial with large $n$ can be approximated by a normal distribution.

Modelling with the binomial

Setting up a binomial problem means answering three questions:

1. What is a trial? One free throw, one component tested, one person surveyed.
2. What counts as a success? Define it explicitly — and note it need not be a good thing. "Defective" is a perfectly valid success.
3. Are the four conditions met? State them if the question asks you to justify the model.

A basketball player scores 70% of free throws and takes 12. Find the probability of scoring exactly 9.

A trial is one throw; success is scoring; $p = 0.7$ is constant if the throws are independent and the player's form does not change.

$$X \sim B(12, 0.7) \qquad P(X = 9) = \binom{12}{9}(0.7)^9(0.3)^3 = 0.240 \text{ (3 s.f.)}$$