Using Venn diagrams for conditional probability - Higher
Venn diagramNamed after John Venn who created the term, it’s a way of identifying mathematical relationships between different groups or sets of things. can be useful for organising information about frequencies and probabilityThe extent to which something is likely to be the case., which can then be used to solve conditional probability problems.
Example
- 90 pupils were asked whether they owned a laptop or a tablet device.
- 52 said they owned a laptop.
- 45 said they owned a tablet.
- 23 said they owned both.
Find the probability that a pupil chosen at random owns a laptop, given that they own exactly one device.
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The Venn diagram shows there are 29 + 22 = 51 pupils who own exactly one device (this becomes the denominatorThe bottom part of a fraction. For ⅝, the denominator is 8, which represents 'eighths'. of the conditional probability). Out of these 51 pupils, 29 own a laptop. Therefore, the probability that a pupil chosen at random owns a laptop, given that they own exactly one device = \(\frac{29}{51}\).
Question
125 pupils were asked about their pets. 61 pupils said they had a cat and 68 pupils said they had a dog. 23 pupils said they had neither a cat nor a dog.
Show this information on a Venn diagram. Find the probability that a pupil chosen at random has a cat, given that they have a dog.
The total number of pupils with either a cat or dog is \(125 - 23 = 102\).
The total number of pupils with both a cat and a dog = (61 + 68) – 102 = 27
The number of pupils with a cat is \(61 – 27 = 34\)
The number of pupils with a dog is \(68 - 27 = 41\)
This information is shown on the Venn diagram.
The Venn diagram shows there are \(27 + 41 = 68\) pupils who have a dog (this becomes the denominator of the conditional probability). Out of these 68 pupils, 27 have a cat. Therefore, the probability that a pupil chosen at random has a cat, given that they have a dog = \(\frac{27}{68}\).