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CIE A-Level Maths Study Notes

4.4.2 Binomial Distribution

The Binomial Distribution is a cornerstone concept in the field of probability and statistics, particularly pivotal for students. This distribution helps in understanding phenomena where the outcome of an experiment or event can be classified into two distinct categories: success and failure.

Binomial Distribution: An Overview

At the heart of the binomial distribution, denoted as B(n, p), are certain conditions:

  • The number of trials, n, is predetermined and finite.
  • Each trial is independent, meaning the outcome of one trial does not influence another.
  • The probability of success, p, is consistent across trials.
  • Only two outcomes are possible: success or failure.
binomial distribution graph

Image courtesy of technologynetworks

The Binomial Formula

The probability of exactly x successes in n trials in a binomial distribution is given by:

P(X=x)=(nx)px(1p)nxP(X = x) = \binom{n}{x} p^x (1-p)^{n-x}

Here, (nx)\binom{n}{x} represents the binomial coefficient and is computed as n!x!(nx)!\frac{n!}{x!(n-x)!}.

Example Problems

Example 1: Quality Control in Manufacturing

Suppose a factory's production line has a 5% defect rate. If we randomly select 10 items for quality control, what is the probability that exactly 2 of them are defective?


  • Number of trials nn: 10
  • Probability of a defective item pp: 0.05
  • Desired number of defective items xx: 2

Using the binomial formula:

P(X=2)=(102)×0.052×0.958P(X = 2) = \binom{10}{2} \times 0.05^2 \times 0.95^8

The binomial coefficient (\binom{10}{2}) is calculated as:

10!2!(102)!=10×92×1=45\frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45


P(X=2)=45×0.052×0.958P(X = 2) = 45 \times 0.05^2 \times 0.95^8

45×0.0025×0.6634\approx 45 \times 0.0025 \times 0.6634

0.0746\approx 0.0746

So, there's about a 7.46% chance of finding exactly 2 defective items.

graph of binomial distribution

Example 2: Sports Probability

Consider a football player with a 70% chance of scoring a goal on each penalty kick. If they take 5 penalty kicks, what is the probability of scoring exactly 3 goals?


  • Number of penalty kicks nn: 5
  • Probability of scoring pp: 0.70
  • Desired number of goals xx: 3

Applying the binomial formula:

P(X=3)=(53)×0.703×0.302P(X = 3) = \binom{5}{3} \times 0.70^3 \times 0.30^2

The binomial coefficient (53)\binom{5}{3} is:

5!3!(53)!=5×42×1=10\frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10


P(X=3)=10×0.703×0.302P(X = 3) = 10 \times 0.70^3 \times 0.30^2

10×0.343×0.09\approx 10 \times 0.343 \times 0.09

0.3087\approx 0.3087

Hence, the player has approximately a 30.87% chance of scoring exactly 3 goals out of 5 kicks.

graph of binomial distribution
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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