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 seensee 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... exactly 3; less than 3, and; 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:/**/ Expectation and variance If X ~ bin(n, p) Then: E(X) = n × p Var(X) = n × p × q n = number of trialsp = probability of a successq = probability of a failure = 1 − pExample: If X ~ bin(26, 0.2) Then: E(X) = 26 × 0.2 = 5.2 Var(X) = 26 × 0.2 × 0.8 = 4.16