Probability with Venn Diagrams

Lessons

Senior 3 · Mathematics

Probability with Venn Diagrams

3 lessons

Lesson 2 of 3

Set concepts and counting outcomes with Venn diagrams

25 min

Two circles, four regions

In lesson 1 a month was chosen at random, with $A$ "the month has 31 days" and $B$ "the name begins with J". January and July belong to both. Drawn as a Venn diagram, the two events cut the sample space into four regions, and every one of the twelve months falls into exactly one of them.

Each region is an event in its own right, with its own probability: count the months in it and divide by 12. The work of this lesson is to name the regions correctly in words and in symbols, because an exam question gives you the words ("both", "at least one", "neither", "only") and expects you to find the right region.

From set language to event language

Symbol	The event happens when…	Read as	Region on the diagram
$A \cap B$	both $A$ and $B$ happen	"A and B"	the overlap
$A \cup B$	$A$ or $B$ or both happen	"A or B", "at least one"	everything inside the circles
$A^{'}$	$A$ does not happen	"not A"	everything outside circle $A$
$A \cap B'$	$A$ happens but $B$ does not	"A only"	$A$ without the overlap
$(A \cup B)' = A' \cap B'$	neither event happens	"neither A nor B"	outside both circles
$(A \cap B') \cup (A' \cap B)$	exactly one of them happens	"exactly one"	the two "only" regions

The fifth row is one of De Morgan's laws from unit 1: "not (A or B)" is the same event as "not A and not B".

Definition

Two events are mutually exclusive if they cannot happen together: they have no outcome in common, so $A \cap B = \varnothing$ and $P(A \cap B) = 0$ . On a Venn diagram their circles do not overlap.

Example: for a number chosen from 1 to 20, "the number is a multiple of 7" $= \{7, 14\}$ and "the number is a multiple of 9" $= \{9, 18\}$ are mutually exclusive; "the number is prime" and "the number is even" are not, because 2 is both.

Worked example 1: the months, region by region

Using the diagram, find (a) $P(A \cap B)$ , (b) $P(A \cup B)$ , (c) $P(A \text{ only})$ , (d) $P(\text{neither})$ , (e) $P(\text{exactly one})$ .

Solution $n (S) = 12$ .

(a) The overlap holds Jan and Jul: $\frac{2}{12} = \frac{1}{6}$ .

(b) The circles hold $5 + 2 + 1 = 8$ months: $\frac{8}{12} = \frac{2}{3}$ . Unit 1's formula gives the same count: $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 7 + 3 - 2 = 8$ .

(d) Feb, Apr, Sep, Nov: $\frac{4}{12} = \frac{1}{3}$ . Or by the complement: $1 - \frac{2}{3} = \frac{1}{3}$ .

(e) The two "only" regions: $\frac{5 + 1}{12} = \frac{1}{2}$ .

Important

Dividing unit 1's counting formula by $n (S)$ turns it into a probability rule:

n(A \cup B) = n(A) + n(B) - n(A \cap B)

becomes

P(A \cup B) = P(A) + P(B) - P(A \cap B).

The overlap is subtracted once because it was counted twice, once inside $A$ and once inside $B$ . If $A$ and $B$ are mutually exclusive there is no overlap, and $P(A \cup B) = P(A) + P(B)$ . Lesson 3 uses this addition rule when the information comes as probabilities rather than counts.

Worked example 2: a coffee and banana cooperative

A cooperative has 50 farmers (illustrative data): 32 grow coffee ( $C$ ), 21 grow bananas ( $B$ ) and 12 grow both. One farmer is chosen at random to attend a training day. Find the probability that the farmer grows (a) coffee only, (b) coffee or bananas, (c) neither crop, (d) exactly one of the two crops.

Solution Fill the diagram from the inside out, as in unit 1: both = 12; coffee only $= 32 - 12 = 20$ ; bananas only $= 21 - 12 = 9$ ; neither $= 50 - (20 + 12 + 9) = 9$ .

(a) $P(C \cap B') = \frac{20}{50} = \frac{2}{5}$ .

(b) $P(C \cup B) = \frac{20 + 12 + 9}{50} = \frac{41}{50}$ . By the rule: $\frac{32}{50} + \frac{21}{50} - \frac{12}{50} = \frac{41}{50}$ .

(d) $\frac{20 + 9}{50} = \frac{29}{50}$ .

Three events: eight regions

With three events the rectangle is cut into eight regions. Here a number is chosen at random from 1 to 30, with $A$ "a multiple of 2", $B$ "a multiple of 3" and $C$ "a multiple of 5". Each number is placed in its region: 30 is in all three circles, 6, 12, 18 and 24 are in $A$ and $B$ but not $C$ , and so on. Eight numbers (1 and the primes from 7 to 29) are multiples of none of them.

Worked example 3: probabilities from three circles

A number is chosen at random from 1 to 30. Find the probability that it is a multiple of (a) none of 2, 3 and 5; (b) exactly one of them; (c) at least two of them; (d) 2 and 3 but not 5; (e) 2 or 3.

Solution $n (S) = 30$ . Read the counts from the diagram.

(a) Outside all three: 8 numbers, so $\frac{8}{30} = \frac{4}{15}$ .

(b) The three "only" regions: $8 + 4 + 2 = 14$ , so $\frac{14}{30} = \frac{7}{15}$ .

(c) The three "two only" regions and the centre: $4 + 2 + 1 + 1 = 8$ , so $\frac{8}{30} = \frac{4}{15}$ . Check: $8 + 14 + 8 = 30$ ✓.

(d) $A \cap B \cap C' = \{6, 12, 18, 24\}$ , so $\frac{4}{30} = \frac{2}{15}$ .

(e) $n(A \cup B) = 15 + 10 - 5 = 20$ , since the multiples of 6 (6, 12, 18, 24, 30) are in both. So $P(A \cup B) = \frac{20}{30} = \frac{2}{3}$ .

Worked example 4: which pairs are mutually exclusive?

A number is chosen from 1 to 30. Decide whether each pair of events is mutually exclusive: (a) "a multiple of 7" and "a multiple of 8"; (b) "a prime" and "an even number"; (c) "a square number" and "a multiple of 5".

Solution

(a) Multiples of 7: $\{7, 14, 21, 28\}$ ; multiples of 8: $\{8, 16, 24\}$ . No common element (the first common multiple, 56, is above 30), so they are mutually exclusive, and $P(\text{7 or 8}) = \frac{4 + 3}{30} = \frac{7}{30}$ .

(b) 2 is prime and even, so they are not mutually exclusive.

Warning

"A" is not "A only". $P (A)$ counts the whole circle, overlap included; "A only" leaves the overlap out. In Worked example 2, $P(\text{coffee}) = \frac{32}{50}$ but $P(\text{coffee only}) = \frac{20}{50}$ . And "or" in probability means "at least one", which includes both: adding $\frac{32}{50} + \frac{21}{50} = \frac{53}{50}$ without removing the overlap gives a probability greater than 1, which is impossible.

Tip

When the regions of a diagram are complete, add them all: the total must be $n (S)$ , or 1 if they are probabilities. It takes ten seconds and catches most filling errors before they spoil every answer that follows.

Khan Academy: Probability with playing cards and Venn diagrams. Watch how the outcomes in the overlap are counted once, not twice, when finding the probability of one event or another.

✦

Practice

A number is chosen at random from 1 to 12. Let $A$ be "even" and $B$ be "a multiple of 3". Draw the Venn diagram and find $P(A \cap B)$ , $P(A \cup B)$ and $P(\text{neither})$ .
Write in set notation: (a) $A$ happens but $B$ does not; (b) neither $A$ nor $B$ happens; (c) at least one of $A$ and $B$ happens.
Of 60 S3 students, 35 study French, 28 study Kiswahili and 10 study both. One student is chosen at random. Find the probability that the student studies (a) neither language, (b) exactly one language, (c) Kiswahili only.
One number is chosen from 1 to 50. Are the events mutually exclusive? (a) "a multiple of 9" and "a multiple of 11"; (b) "a square number" and "a multiple of 7".
Use the three-circle diagram of the numbers 1 to 30 to find the probability that the number chosen is a multiple of 3 or 5. Check your answer with the formula.
In a sample space of 50 equally likely outcomes, $n (A) = 20$ , $n (B) = 25$ and $P(A \cup B) = \frac{37}{50}$ . Find $P(A \cap B)$ .
In a class of 30 pupils, the probability that a pupil chosen at random plays football is $\frac{1}{2}$ , the probability that the pupil sings in the choir is $\frac{1}{3}$ , and 6 pupils do neither. Find the probability that a pupil chosen at random does both.

Show answersHide answers

$A = \{2, 4, 6, 8, 10, 12\}$ , $B = \{3, 6, 9, 12\}$ , $A \cap B = \{6, 12\}$ . $P(A \cap B) = \frac{2}{12} = \frac{1}{6}$ ; $P(A \cup B) = \frac{6 + 4 - 2}{12} = \frac{8}{12} = \frac{2}{3}$ ; neither $= \{1, 5, 7, 11\}$ , so $\frac{4}{12} = \frac{1}{3}$ .
(a) $A \cap B'$ ; (b) $(A \cup B)'$ , or $A' \cap B'$ ; (c) $A \cup B$ .
French only 25, Kiswahili only 18, both 10, neither $60 - 53 = 7$ . (a) $\frac{7}{60}$ ; (b) $\frac{25 + 18}{60} = \frac{43}{60}$ ; (c) $\frac{18}{60} = \frac{3}{10}$ .
(a) Yes: the first common multiple of 9 and 11 is 99, above 50. (b) No: 49 is a square and a multiple of 7.
Regions of $B \cup C$ : $4 + 4 + 1 + 1 + 2 + 2 = 14$ , so $\frac{14}{30} = \frac{7}{15}$ . Formula: $n(B \cup C) = 10 + 6 - 2 = 14$ (15 and 30 are in both).
$n(A \cup B) = 37$ , so $n(A \cap B) = 20 + 25 - 37 = 8$ and $P(A \cap B) = \frac{8}{50} = \frac{4}{25}$ .
Football 15, choir 10, at least one $30 - 6 = 24$ . Both $= 15 + 10 - 24 = 1$ , so the probability is $\frac{1}{30}$ .

Sources and credits

Video: Probability with playing cards and Venn diagrams, Khan Academy, embedded from YouTube.
Written by Ganzaa. The cooperative and class data are illustrative, and the diagrams, worked examples and practice items were written by Ganzaa.

1Two circles, four regions

2From set language to event language

3Three events: eight regions

Two circles, four regions

From set language to event language

Three events: eight regions