Venn diagrams and notation — Expectation

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Venn diagrams and notation — the key ideas

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Step 1 · Sets and notation

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A set is a collection of outcomes. $P(A)$ is the probability of $A$. $A'$ means not $A$, so $P(A') = 1 - P(A)$. $A \cap B$ means both $A$ and $B$. $A \cup B$ means $A$ or $B$ or both. The whole rectangle is the sample space.

Step 2 · Filling in a Venn diagram

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Start in the middle (the overlap), then work outwards: set total minus overlap gives the 'only' regions. Finally, anything left over goes in the 'neither' region outside both circles. All the numbers must add to the total.

Step 3 · The addition rule

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$P(A \cup B) = P(A) + P(B) - P(A \cap B)$. We subtract the overlap because it has been counted twice. If $A$ and $B$ cannot happen together ($P(A \cap B) = 0$) they are mutually exclusive and the rule is simply $P(A) + P(B)$.

Step 4 · Equally likely outcomes and Fermat's method

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When outcomes are equally likely, $P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$. Fermat's method shares a prize fairly when a game is stopped early: list all the equally likely ways the remaining games could go, find the fraction in which each player wins, and split the prize in that ratio.

Worked example

In a class of 30, 18 study Maths (M), 13 study Physics (P) and 5 study both. Draw a Venn diagram and find $P(M \cup P)$.

Both = 5. Maths only = 18 − 5 = 13. Physics only = 13 − 5 = 8. Neither = 30 − (13 + 5 + 8) = 4.

$P(M \cup P) = \dfrac{13 + 5 + 8}{30} = \dfrac{26}{30} = \dfrac{13}{15}$. Check with the addition rule: $\frac{18}{30} + \frac{13}{30} - \frac{5}{30} = \frac{26}{30}$ ✓.

Test yourself

Past-paper style question

80 people were asked how they travel to work. 45 use the bus, 30 use the train and 12 use both.

(a) Draw a Venn diagram to show this information.
(b) A person is chosen at random. Find the probability that they use the bus or the train (or both).
(c) Find the probability that they use neither. [5 marks]

Show the answer

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(a) Both = 12; bus only = 45 − 12 = 33; train only = 30 − 12 = 18; neither = 80 − (33 + 12 + 18) = 17. (3 marks)

(b) $P(B \cup T) = \dfrac{33 + 12 + 18}{80} = \dfrac{63}{80}$. (1 mark)

(c) $P(\text{neither}) = \dfrac{17}{80}$. (1 mark)

Practice

Venn diagrams and notation worksheet

Five short questions. Work through them, then reveal the mark scheme to check.

$P(A) = 0.4$, $P(B) = 0.5$ and $P(A \cap B) = 0.2$. Find $P(A \cup B)$.
A fair six-sided dice is rolled. Find the probability of a prime number.
In a class of 25, 15 play football, 10 play tennis and 3 play both. How many play neither?
$P(A \cup B) = 0.8$, $P(A) = 0.5$ and $P(B) = 0.6$. Find $P(A \cap B)$.
Two players play a first-to-3 match for a £60 prize. Player A leads 2–1 when the match has to stop. Use Fermat's method (list the results of 2 more games) to share the prize fairly.

Mark scheme

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$0.4 + 0.5 - 0.2 = $ 0.7.
Primes: 2, 3, 5 → 3/6 = 1/2.
Football only 12, both 3, tennis only 7, so 12+3+7 = 22 play at least one. Neither = 25 − 22 = 3.
$P(A \cap B) = 0.5 + 0.6 - 0.8 = $ 0.3.
Play 2 more games: AA, AB, BA, BB (equally likely). A wins the match in AA, AB and BA (3 of 4); B only in BB. A: ¾ × £60 = £45; B: ¼ × £60 = £15.

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