Random Variables

Lessons

Senior 6 · Mathematics

Random Variables

5 lessons

Lesson 3 of 5

Binomial and Poisson distributions

40 min

Two models that appear everywhere

Many random counts come from one of two situations. In the first, a fixed number of independent attempts each succeed or fail with the same probability: seeds that germinate or not, questions answered right or wrong, items that pass or fail an inspection. That count follows a binomial distribution. In the second, events happen one at a time, at random moments, at a steady average rate: customers arriving at a counter, calls reaching a switchboard, typing errors on a page. That count follows a Poisson distribution, named after the French mathematician Siméon Denis Poisson (1781–1840).

Definition

A Bernoulli trial is an experiment with two outcomes, "success" (probability $p$ ) and "failure" (probability $q = 1 - p$ ).

Let $X$ be the number of successes in $n$ Bernoulli trials that are independent and all have the same $p$ . Then $X$ has the binomial distribution $B (n, p)$ , written $X \sim B(n, p)$ :

P(X = k) = \binom{n}{k} p^k q^{n - k} , \qquad k = 0, 1, \dots, n .

Example: the number of sixes in 5 rolls of a fair die is $B\left(5, \frac{1}{6}\right)$ .

Important

Why the formula holds. Any particular order of $k$ successes and $n - k$ failures has probability $p^kq^{n - k}$ (independence), and there are $\binom{n}{k}$ ways to choose which $k$ trials are the successes. By the binomial theorem, the probabilities add up to $(p + q)^{n} = 1$ .

Mean and variance. If $X \sim B(n, p)$ , then

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

( $X$ is a sum of $n$ independent Bernoulli variables, each with mean $p$ and variance $p - p^{2} = p q$ ; means and, for independent variables, variances add.)

Worked example 1: germinating seeds

Suppose (illustratively) that each bean seed in a batch germinates with probability 0.8, independently. A farmer plants 10 seeds. Find the probability that exactly 8 germinate, the probability that at least 9 germinate, and the mean and standard deviation of the number that germinate.

Solution. $X \sim B(10, 0.8)$ .

P(X = 8) = \binom{10}{8}(0.8)^8(0.2)^2 = 45 \times 0.16777 \times 0.04 \approx 0.302 .

P(X \ge 9) = \binom{10}{9}(0.8)^9(0.2) + (0.8)^{10} \approx 0.2684 + 0.1074 = 0.376 .

$E(X) = 10 \times 0.8 = 8$ , $\operatorname{Var}(X) = 10 \times 0.8 \times 0.2 = 1.6$ , $\sigma \approx 1.26$ .

Worked example 2: guessing in a test

A pupil guesses all 10 questions of a multiple-choice test, each with 4 options. Find the probability of no correct answers, and of at least 5.

Solution. $X \sim B(10, 0.25)$ . $P(X = 0) = 0.75^{10} \approx 0.0563$ . For at least 5, add $P (X = k)$ for $k = 5, \dots, 10$ : $P(X \ge 5) \approx 0.0584 + 0.0162 + 0.0031 + 0.0004 + 0.0000 + 0.0000 \approx 0.0781$ . Guessing gives on average $n p = 2.5$ correct answers, and passing with half marks happens fewer than 8 times in 100.

Definition

Let $X$ count the events in a fixed interval (of time, length, area …) when events occur singly, independently of each other, and at a constant average rate. If the mean number per interval is $\lambda \gt 0$ , $X$ has the Poisson distribution $\operatorname{Po}(\lambda)$ :

P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} , \qquad k = 0, 1, 2, \dots

with $E(X) = \operatorname{Var}(X) = \lambda$ . The probabilities add up to 1 because $\sum_{k \ge 0} \frac{\lambda^k}{k!} = e^{\lambda}$ . Example: if a page has on average 1.5 typing errors, the number of errors on a page is modelled by $\operatorname{Po}(1.5)$ .

Worked example 3: customers at an agent

In an illustrative model, customers reach a mobile-money agent at an average rate of 3 per 10 minutes. Find the probability that (a) no customer arrives in a 10-minute period; (b) at least 2 arrive in 10 minutes; (c) exactly 4 arrive in 20 minutes.

Solution. (a) $X \sim \operatorname{Po}(3)$ : $P(X = 0) = e^{-3} \approx 0.0498$ .

(b) $P(X \ge 2) = 1 - P(X = 0) - P(X = 1) = 1 - e^{-3}(1 + 3) = 1 - 4e^{-3} \approx 0.801$ .

(c) The rate scales with the interval: in 20 minutes the mean is $\lambda = 6$ . $P(Y = 4) = \frac{e^{-6}6^4}{4!} = 54e^{-6} \approx 0.134$ .

Important

Poisson approximation to the binomial. If $n$ is large and $p$ is small, then $B (n, p)$ is close to $\operatorname{Po}(\lambda)$ with $\lambda = np$ . A common working rule is $n \ge 50$ and $np \le 5$ .

Reason. With $p = \frac{\lambda}{n}$ , $\binom{n}{k}p^k(1 - p)^{n - k} = \frac{n(n - 1)\cdots(n - k + 1)}{n^k} \cdot \frac{\lambda^k}{k!} \cdot \left(1 - \frac{\lambda}{n}\right)^{n}\left(1 - \frac{\lambda}{n}\right)^{-k}$ . As $n \to \infty$ , the first and last factors tend to 1 and $\left(1 - \frac{\lambda}{n}\right)^n \to e^{-\lambda}$ .

When

n

is large and

p

small, the binomial distribution (blue) and the Poisson distribution with the same mean (orange) are almost identical. Bars drawn from computed probabilities.

Worked example 4: defective items

A machine makes items of which 1% are defective, independently. Estimate the probability that a box of 200 items contains at most 2 defective ones.

Solution. $X \sim B(200, 0.01)$ with $n$ large and $p$ small, so use $\operatorname{Po}(2)$ :

P(X \le 2) \approx e^{-2}\left(1 + 2 + \frac{2^2}{2!}\right) = 5e^{-2} \approx 0.677 .

The exact binomial value is also 0.677 to three decimal places, as the figure shows.

Warning

Check the conditions before using a model. Drawing 5 cards from 52 without replacement is not binomial: $p$ changes after every draw. Arrivals that come in groups (a bus unloading passengers) are not Poisson: events must occur singly.

Match $\lambda$ to the interval asked about. A rate of 3 per 10 minutes is a mean of 6 per 20 minutes and 0.3 per minute.

Tip

"At least one" is almost always quickest through the complement: $P(X \ge 1) = 1 - P(X = 0)$ . For $B (n, p)$ this is $1 - q^{n}$ ; for $\operatorname{Po}(\lambda)$ it is $1 - e^{-\lambda}$ .

Khan Academy: Poisson process 1 (about 11 min). Watch how counting cars passing a point is first modelled as a binomial with many small time slots, and how letting the number of slots grow leads to the Poisson formula, as in the approximation above.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks you to reason.

$X \sim B(6, 0.5)$ . Find $P (X = 3)$ and $P(X \le 1)$ .
$X \sim B(20, 0.3)$ . Find the mean, variance and standard deviation of $X$ .
$X \sim \operatorname{Po}(4)$ . Find $P (X = 2)$ and $P(X \ge 1)$ .
A player scores 70% of her free throws, independently. Find the probability that she scores at least 7 of her next 8.
Typing errors occur at an average of 1.5 per page (illustrative). Find the probability that a 4-page letter contains at most 3 errors.
$X \sim B(100, 0.02)$ . Find $P (X = 3)$ exactly and with the Poisson approximation, and compare.
Find the smallest number of rolls of a fair die for which the probability of at least one six is more than 0.9.

Show answersHide answers

$P(X = 3) = \frac{20}{64} = 0.3125$ ; $P(X \le 1) = \frac{1 + 6}{64} \approx 0.109$ .
$E (X) = 6$ , $\operatorname{Var}(X) = 4.2$ , $\sigma \approx 2.05$ .
$P(X = 2) = 8e^{-4} \approx 0.147$ ; $P(X \ge 1) = 1 - e^{-4} \approx 0.982$ .
$8(0.7)^7(0.3) + (0.7)^8 \approx 0.198 + 0.058 = 0.255$ .
$\lambda = 6$ : $e^{-6}(1 + 6 + 18 + 36) = 61e^{-6} \approx 0.151$ .
Exact $\binom{100}{3}(0.02)^3(0.98)^{97} \approx 0.182$ ; Poisson $\frac{e^{-2}2^3}{3!} \approx 0.180$ : close, with $n p = 2$ .
$1 - \left(\frac{5}{6}\right)^n \gt 0.9 \iff n \gt \frac{\ln 0.1}{\ln\frac{5}{6}} \approx 12.6$ , so $n = 13$ (check: $1 - \left(\frac{5}{6}\right)^{13} \approx 0.907$ , while 12 rolls give 0.888).

Sources and credits

Written by Ganzaa: the explanations, the derivations, the chart (drawn from computed probabilities), the worked examples, the practice items and the quiz questions. The germination rate, the customer rate and the error rate are illustrative, not measured data.
Video: Poisson process 1, Khan Academy, embedded from YouTube.

1Two models that appear everywhere

Two models that appear everywhere