Suppose that an experiment consists of n identical and independent trials. For example flipping a coin over and over again n times. For each trial there are 2 outcomes.

'Success' - which is given probability p

'Failure' - which is given probability q where q = 1 − p

Then if X = the number of successes, we say that X has a binomial distribution.

We write:

This is sometimes written as: X ~ B(n, p)

If our random variable follows a binomial distribution, then the associated probabilities are calculated using the following formula:

Note: If you haven't seen

see the section on Combinations.

gives us the number of ways of choosing r objects from n and is calculated by:

You may also have a button on your calculator that will do all that for you.

Let's see this in action...

Example:

The probability that sixth formers know what the first prime number is, 0.35.

Find the probability that in a sample of 14 sixth formers, the number who know is...

1. exactly 3;
2. less than 3, and;
3. greater than 1.

If we let X be the number of successes, our distribution is given by:

Using the formula:

Therefore:

Therefore:

To get P(X > 1), we could calculate this by working out: P(X = 2) + P(X = 3) + P(X = 4)... + P(X = 14), taking a very long time and getting extremely bored! Instead, we use the fact that these distributions sum to 1 (they are exhaustive!)

Therefore:

Question:

See if you can work out the following probabilities if:

If X ~ bin(n, p)

Then:

• E(X) = n × p
• Var(X) = n × p × q

n = number of trials

p = probability of a success

q = probability of a failure = 1 − p

Example:

If X ~ bin(26, 0.2)

Then: E(X) = 26 × 0.2 = 5.2

Var(X) = 26 × 0.2 × 0.8 = 4.16