Basic probability rules: mutually exclusive and inclusive events – Elementary Probability (Senior 6 Basic Mathematics) | Ganzaa

Elementary Probability

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Senior 6 · Basic Mathematics

Elementary Probability

5 lessons

Lesson 2 of 5

Basic probability rules: mutually exclusive and inclusive events

25 min

"Or" in probability

In an S6 class of 40 (illustrative figures), 18 students take Economics, 15 take Geography and 7 take both. A student is chosen at random to represent the class. What is the probability that the student takes Economics or Geography? Adding $\frac{18}{40} + \frac{15}{40}$ gives $\frac{33}{40}$ , which is too big: the 7 students who take both have been counted twice. In Unit 2 you corrected the same double count with $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ . This lesson turns that counting rule into a probability rule.

Definition

$A \cup B$ (" $A$ or $B$ ") is the event that $A$ happens, or $B$ happens, or both. $A \cap B$ (" $A$ and $B$ ") is the event that both happen.
Events $A$ and $B$ are mutually exclusive if they cannot happen together: $A \cap B = \varnothing$ , so $P(A \cap B) = 0$ . Example: drawing a king and drawing a queen with one card.
Events that can happen together are inclusive (not mutually exclusive): $P(A \cap B) > 0$ . Example: drawing a king and drawing a heart (the king of hearts is both).
Events are exhaustive if together they cover the whole sample space: $A \cup B = S$ . An event and its complement are always mutually exclusive and exhaustive.

The addition rule

For any two events $A$ and $B$ in a sample space of equally likely outcomes,

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

and dividing every term by $n (S)$ gives the general addition rule:

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

The rule holds for all events, not only equally likely outcomes. If $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$ and the rule simplifies to the special addition rule:

P(A \cup B) = P(A) + P(B)

It extends to any number of mutually exclusive events: $P(A \cup B \cup C) = P(A) + P(B) + P(C)$ .

Worked example

Worked example 1: inclusive events. For the class of 40 above, let $E$ = "takes Economics" and $G$ = "takes Geography". Find (a) $P(E \cup G)$ ; (b) $P(\text{neither})$ ; (c) $P(\text{exactly one of the two})$ .

(a) $P(E \cup G) = \frac{18}{40} + \frac{15}{40} - \frac{7}{40} = \frac{26}{40} = \frac{13}{20}$

(b) "Neither" is the complement of $E \cup G$ : $P = 1 - \frac{13}{20} = \frac{7}{20}$ . The Venn diagram shows the 14 students outside both circles: $\frac{14}{40} = \frac{7}{20}$ .

Worked example

Worked example 2: cards. One card is drawn from a standard pack of 52. Find (a) $P(\text{king or heart})$ ; (b) $P(\text{king or queen})$ .

(a) Inclusive: the king of hearts is both.

P(K \cup H) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}

(b) Mutually exclusive: no card is both a king and a queen.

P(K \cup Q) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}

Worked example

Worked example 3: working from probabilities. $P (A) = 0.5$ , $P (B) = 0.4$ and $P(A \cup B) = 0.7$ . Find $P(A \cap B)$ and decide whether $A$ and $B$ are mutually exclusive.

Rearranging the addition rule, $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.7 = 0.2$ .

Since $P(A \cap B) = 0.2 \ne 0$ , the events are not mutually exclusive.

Warning

Check for overlap before adding. $P (A) + P (B)$ is only correct when the events cannot happen together. If $P (A) + P (B) > 1$ , the events must overlap, and the overlap must be subtracted.

Tip

Fill in a Venn diagram from the inside out: put $n(A \cap B)$ in the overlap first, then " $A$ only" and " $B$ only", then the region outside. Every probability question about two events can then be read off the diagram.

Khan Academy: "Addition rule for probability". Watch how the overlap is subtracted, and compare with Worked example 2.

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Practice

Which pairs are mutually exclusive when one card is drawn? (a) a club and a red card; (b) an ace and a spade; (c) a picture card and a 7.
Two fair dice are rolled. $A$ = "the total is 7", $B$ = "a double". Show that $A$ and $B$ are mutually exclusive and find $P(A \cup B)$ .
One card is drawn from a pack of 52. Find $P(\text{red or ace})$ .
$P (A) = 0.35$ , $P (B) = 0.45$ , $P(A \cap B) = 0.15$ . Find (a) $P(A \cup B)$ ; (b) $P(A' \cap B')$ ; (c) $P(A \cap B')$ .
In a survey of 60 households in a village (illustrative), 34 have a radio, 25 a mobile phone with internet, and 10 both. One household is chosen at random. Find the probability that it has (a) at least one of these; (b) neither.
$A$ and $B$ are mutually exclusive with $P (A) = 0.3$ and $P(A \cup B) = 0.75$ . Find $P (B)$ .
Explain why two events with $P (A) = 0.6$ and $P (B) = 0.7$ cannot be mutually exclusive.

Show answersHide answers

(a) yes (b) no: the ace of spades (c) yes.
A double has equal scores, so its total is even and can never be 7. $P(A \cup B) = \frac{6}{36} + \frac{6}{36} = \frac{12}{36} = \frac{1}{3}$ .
$\frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$ .
(a) $0.35 + 0.45 - 0.15 = 0.65$ (b) $1 - 0.65 = 0.35$ (c) $0.35 - 0.15 = 0.20$ .
(a) $\frac{34 + 25 - 10}{60} = \frac{49}{60}$ (b) $\frac{11}{60}$ .
$P (B) = 0.75 - 0.3 = 0.45$ .
If they were, $P(A \cup B) = 0.6 + 0.7 = 1.3 > 1$ , which is impossible.

Important

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ always. For mutually exclusive events $P(A \cap B) = 0$ , so $P(A \cup B) = P(A) + P(B)$ .

Sources and credits

Text, Venn diagrams, worked examples and exercises: Written by Ganzaa. Class and survey figures are illustrative.

Video: Addition rule for probability, Khan Academy, embedded from YouTube.

1"Or" in probability

2The addition rule

"Or" in probability

The addition rule