Bernoulli trials and the binomial distribution

What you will be able to do

• recognise Bernoulli trials and decide whether a situation is binomial;
• use $P(X = k) = \binom{n}{k}p^k q^{n-k}$;
• use the mean $np$ and the variance $npq$.

Ten free throws

A basketball player takes free throws. Each throw either goes in or it does not, and we count only that. One throw is a Bernoulli trial: a trial with exactly two outcomes, called success and failure. The chance of success is $p$, and the chance of failure is $q = 1 - p$.

Bernoulli trials are $n$ such trials that are independent, with the same $p$ every time. If a player takes $5$ throws and goes in with probability $0.6$ each time, how many throws will go in? The count $X$ is random. It can be $0$ to $5$. The question is how likely each count is.

The formula

Take $3$ trials and ask for exactly $2$ successes. The outcomes are SSF, SFS and FSS. Each has two successes and one failure, so each has probability $p \times p \times q$. There are $3$ of them, so $P(X = 2) = 3p^2q$.

For $n$ trials, the number of outcomes with exactly $k$ successes is the number of ways to choose the $k$ places, which is $\binom{n}{k}$. Each has probability $p^k q^{n-k}$. So

We say that $X$ has the binomial distribution with parameters $n$ and $p$, and write $X \sim B(n, p)$. IB and AP write $X \sim \text{Bin}(n, p)$ and use the calculator command binompdf. The probabilities are the terms of $(q + p)^n$, so they add up to $1$. The binomial theorem of the earlier chapter is the reason.

Four conditions

A situation is binomial when all four of these hold.

1. There is a fixed number $n$ of trials.
2. Each trial has two outcomes, success and failure.
3. The probability of success is the same $p$ on every trial.
4. The trials are independent.

Tossing a coin $10$ times passes all four. Drawing cards from a pack without replacement fails the fourth, because each draw changes the pack. Counting the throws needed to get the first six fails the first, because $n$ is not fixed.

Mean and variance

Write $X$ as a sum of $n$ separate indicators, one for each trial. An indicator is $1$ for a success and $0$ for a failure. Each has mean $p$ and variance $p(1 - p) = pq$. The means add, and because the trials are independent, so do the variances:

The standard deviation is $\sqrt{npq}$. The mean $np$ is not always a whole number, and it need not be a value that $X$ can take.

A worked example

A player scores with probability $0.6$ on each of $5$ free throws, independently.

1. Exactly three. $P(X = 3) = \binom{5}{3}(0.6)^3(0.4)^2 = 10 \times 0.216 \times 0.16 = 0.3456$.
2. The whole table. $P(X = 0, 1, 2, 3, 4, 5)$ is $0.01024$, $0.0768$, $0.2304$, $0.3456$, $0.2592$ and $0.07776$. They add up to $1$.
3. Mean and variance. $E(X) = 5 \times 0.6 = 3$ and $\text{Var}(X) = 5 \times 0.6 \times 0.4 = 1.2$. The standard deviation is about $1.095$.

A check: the tallest bar is at $3$, which is also the mean.

Practise with new numbers

Try it in Python

The program builds the whole distribution of $5$ trials with success probability $\dfrac{3}{5}$ from the formula. It then finds the mean and the variance from the table and compares them with $np$ and $npq$.

Common mistakes

• Leaving out $\binom{n}{k}$. The product $p^k q^{n-k}$ is the chance of one particular order only.
• Using the binomial for draws without replacement. The trials are then not independent.
• Mixing up $p$ and $q$. In $P(X = k)$, $p$ goes with the $k$ successes and $q$ with the $n - k$ failures.
• Calling $np$ the most likely value. The mean and the most likely value are close, but they are not the same thing.

Remember these

• $X \sim B(n, p)$ means $P(X = k) = \binom{n}{k}p^k q^{n-k}$ for $k = 0, 1, \dots, n$.
• The four conditions: fixed $n$, two outcomes, the same $p$, independent trials.
• $E(X) = np$ and $\text{Var}(X) = npq$.
• The probabilities add up to $(q + p)^n = 1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Five cards are drawn from a pack without replacement, and $X$ is the number of hearts. Is $X$ exactly binomial?

2. $X \sim B(6, 0.5)$. Find $P(X = 3)$.

3. $X \sim B(10, 0.3)$. Find the mean, the variance and the standard deviation.

4. A fair coin is tossed $6$ times. Find the probability of at least $5$ heads.

5. A binomial variable has mean $12$ and variance $8$. Find $n$ and $p$.

References

• OpenStax, Introductory Statistics, Chapter 4, section on the binomial distribution.
• NCERT, Mathematics Part II for Class XII, Chapter 13, section on Bernoulli trials and the binomial distribution.