Binomial Distribution

AS · S1 · 3 min

The binomial distribution counts successes in a fixed number of independent trials that each succeed with the same probability. It is the model for "how many out of $n$ ", from defective items in a batch to heads in a run of tosses.

The model

Key result

$X \sim B(n, p)$ when there are $n$ independent trials, each with the same probability $p$ of success, and $X$ counts the successes.

P(X = r) = \binom{n}{r}p^r(1 - p)^{n - r}, \qquad r = 0, 1, \ldots, n

E(X) = np, \qquad \operatorname{Var}(X) = np(1 - p)

Check the conditions before using it: fixed $n$ , two outcomes per trial, constant $p$ , independence. Sampling without replacement from a small population breaks the constant- $p$ condition; sampling from a large population is close enough.

Exact probabilities

A machine produces items that are faulty with probability $0.08$ , independently. In a batch of 20, find the probability of (a) exactly 2 faulty; (b) at most 2 faulty; (c) at least 1 faulty.

Solution

$X \sim B(20, 0.08)$ .

(a) $\binom{20}{2}(0.08)^2(0.92)^{18} = 190 \times 0.0064 \times 0.2229 = 0.271$ .

(b) $P(X \leq 2) = 0.92^{20} + 20(0.08)(0.92)^{19} + 0.271 = 0.1887 + 0.3282 + 0.2711 = 0.788$ .

Translating the words

$X \sim B(12, 0.3)$ . Write each of these in terms of $P(X = r)$ values: more than 9; fewer than 3; between 4 and 6 inclusive; at least 10.

Solution

$P(X > 9) = P(10) + P(11) + P(12)$ . $P(X < 3) = P(0) + P(1) + P(2)$ . $P(4 \leq X \leq 6) = P(4) + P(5) + P(6)$ . $P(X \geq 10) = P(X > 9)$ .

Mean and variance

A multiple-choice test has 40 questions each with 5 options. A student guesses every answer. Find the mean and standard deviation of the number of correct answers, and the probability of scoring exactly the mean.

Solution

$X \sim B(40, 0.2)$ : $E(X) = 8$ , $\operatorname{Var}(X) = 40 \times 0.2 \times 0.8 = 6.4$ , $\sigma = 2.53$ .

$P(X = 8) = \binom{40}{8}(0.2)^8(0.8)^{32} = 0.156$ .

Finding n or p

$X \sim B(n, 0.25)$ and $P(X = 0) = 0.0563$ . Find $n$ . Separately, $Y \sim B(8, p)$ with $E(Y) = 2$ ; find $P(Y = 2)$ .

Solution

$0.75^n = 0.0563 \Rightarrow n = \dfrac{\ln 0.0563}{\ln 0.75} = 10.0$ , so $n = 10$ . (In S1, without logs, find $n$ by trial: $0.75^{10} = 0.0563$ .)

$8p = 2 \Rightarrow p = 0.25$ . $P(Y = 2) = \binom{8}{2}(0.25)^2(0.75)^6 = 28 \times 0.0625 \times 0.1780 = 0.311$ .

Binomial inside a binomial

The probability that a seed germinates is $0.7$ . Seeds are sold in packets of 5. A packet is "good" if at least 4 seeds germinate. Find the probability a packet is good, and the probability that in 6 packets exactly 4 are good.

Solution

Per packet, $X \sim B(5, 0.7)$ : $P(X \geq 4) = 5(0.7)^4(0.3) + 0.7^5 = 0.3602 + 0.1681 = 0.528$ .

Packets: $G \sim B(6, 0.528)$ : $P(G = 4) = \binom{6}{4}(0.528)^4(0.472)^2 = 15 \times 0.0777 \times 0.2228 = 0.260$ .

Watch out

"At least 3" means $X \geq 3$ , i.e. $1 - P(X \leq 2)$ . "More than 3" means $X \geq 4$ . Off-by-one errors in the inequality are the most common lost mark.

Exam tip

Write the distribution with its parameters, $X \sim B(20, 0.08)$ , before any calculation. When asked why the binomial is (or is not) suitable, name the condition in context: "each item is faulty independently with the same probability" or "the probability changes because items are not replaced".

Practice

Question

$X \sim B(10, 0.4)$ . Find $P(X = 3)$ , $P(X \leq 1)$ and $P(X \geq 9)$ .
A fair die is rolled 15 times. Find the probability of exactly 3 sixes and the expected number of sixes.
$X \sim B(n, 0.5)$ has variance $5$ . Find $n$ and $P(X = 10)$ .
In a large population $12\%$ are left-handed. Find the probability that in a random sample of 25 at least 2 are left-handed.

Answers

$0.215$ ; $0.6^{10} + 10(0.4)(0.6)^9 = 0.0060 + 0.0403 = 0.0464$ ; $10(0.4)^9(0.6) + 0.4^{10} = 0.00157 + 0.00010 = 0.00168$ .
$\binom{15}{3}\left(\tfrac{1}{6}\right)^3\left(\tfrac{5}{6}\right)^{12} = 0.236$ ; $E = 2.5$ .
$0.25n = 5 \Rightarrow n = 20$ ; $P(X = 10) = \binom{20}{10}(0.5)^{20} = 0.176$ .
$1 - 0.88^{25} - 25(0.12)(0.88)^{24} = 1 - 0.0409 - 0.1395 = 0.820$ .

The model

Practice

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