13.1 Discrete probability distributions and the binomial distribution — Learn

Prerequisites: the probability rules for independent and mutually exclusive events from Section 12.1; the binomial coefficient $\binom{n}{r}$ from Section 4.1.

By the end of this section you can: list the values and probabilities of a discrete random variable; check that a set of probabilities is a valid distribution; recognise the four conditions for a binomial model; write $X \sim B(n, p)$; and work out a single binomial probability from the formula.

1. Discrete random variables

A random variable is a number whose value depends on the outcome of an experiment. A discrete random variable takes values that you can list, one by one. We usually name it with a capital letter, such as $X$.

Example 1. A fair four-sided spinner has faces numbered 1, 2, 3 and 4. The score is a random variable $X$. List the possible values of $X$ and find the probability of each one.

The spinner can land on 1, 2, 3 or 4. Each face is equally likely, so

The values of $X$ are 1, 2, 3 and 4, and each has probability $\tfrac14$.

Example 2. A fair coin is tossed three times. Let $X$ be the number of heads. List the values of $X$ and find the probability of each one.

The number of heads can be 0, 1, 2 or 3. There are eight equally likely sequences:

One sequence gives three heads, three sequences give two heads, three give one head, and one gives no heads. So

2. A probability distribution

The full list of values and their probabilities is the probability distribution of $X$. It is often written as a table. Every probability distribution obeys two rules.

• Each probability is between 0 and 1, because it is a probability.
• The probabilities add up to 1, because the list covers every possible outcome.

Check Example 2:

The same two rules let you find a missing probability. If the probabilities are $0.1$, $0.3$, $k$ and $0.2$, then

3. The binomial model

Many experiments have the same shape. A seed either germinates or it does not. A machine part either passes inspection or it fails. In each case we count the number of times one outcome happens.

A binomial random variable counts the number of successes in a fixed number of trials. Four conditions must all hold.

1. There is a fixed number $n$ of trials.
2. Each trial has two outcomes. We call one a success and the other a failure.
3. The probability of success, $p$, is the same for every trial.
4. The trials are independent. The result of one trial does not change the probability for the next.

When all four conditions hold, we write

The symbol $\sim$ means "is distributed as". The $B$ stands for binomial.

The labelling is a choice. In a seed packet, "success" can be "the seed germinates". Then $p = 0.8$. If you instead call "the seed does not germinate" a success, then $p = 0.2$. State which outcome you call a success before you calculate.

4. The binomial probability formula

For a binomial variable $X \sim B(n, p)$, the probability of exactly $r$ successes is

The three parts of the formula each have a job.

• $\binom{n}{r}$ counts the positions of the $r$ successes among the $n$ trials.
• $p^r$ is the probability that those $r$ trials are successes.
• $(1-p)^{n-r}$ is the probability that the other $n-r$ trials are failures.

Example 3. The random variable $X$ counts the number of sixes in six rolls of a fair die. Find $P(X = 2)$.

There are $n = 6$ rolls, and the probability of a six is $p = \tfrac16$. The complement is $1 - p = \tfrac56$. So

Work out the binomial coefficient and the two powers:

Multiply:

Example 4. From the seed packet, $X \sim B(20, 0.8)$. Find $P(X = 20)$, the probability that every seed germinates.

Every one of the 20 seeds must germinate. There is $\binom{20}{20} = 1$ way for that to happen, and

So all 20 seeds germinate in about 1.15% of packets of this kind.

5. Try it: the shape of a binomial distribution

Predict first. The figure shows the probability distribution of a binomial variable $X \sim B(n, p)$ as a bar chart. One bar has height $P(X = r)$.

Before you move the sliders, predict what happens to the shape when $p$ increases from 0.5 to 0.9. Write down where you think the tallest bar will sit, and whether the chart will stay symmetric.

Test. Move the $p$ slider and watch the bars. Then change $n$ and watch again. Note where the tallest bar sits, and whether the bars are symmetric about it.

Explain. Write two sentences. The first explains how the tallest bar moves when $p$ changes. The second explains why the chart is symmetric when $p = 0.5$ but not when $p = 0.9$.

Watch out. Three mistakes are common.

Using the binomial when the trials are not independent. A bag holds five red and three blue counters. Two counters are taken without replacement. The probability for the second counter depends on the first, so $p$ is not constant. The binomial model does not apply. It does apply when the population is large enough that removing one item changes $p$ very little.

Forgetting the binomial coefficient. The successes can sit in several positions. For $X \sim B(6, \tfrac16)$, the term for two sixes is $15 \times (\tfrac16)^2 (\tfrac56)^4$, not $(\tfrac16)^2 (\tfrac56)^4$. The coefficient $15$ counts the positions.

Using the wrong power for the failure. The powers must add to $n$. For two successes in six trials, the failure power is $6 - 2 = 4$.

Re-check: before using $B(n, p)$, run through the four conditions. After using the formula, check that the two powers add to $n$.

You should now be able to

• list the values of a discrete random variable and their probabilities;
• check that a set of probabilities sums to 1, and find a missing probability;
• test a context against the four binomial conditions;
• write a binomial model as $X \sim B(n, p)$ and define what a success means;
• work out $P(X = r)$ using the binomial formula.