Exhaustive and Mutually Exclusive Events

1 The Probabilities Always Add to One

The fundamental rule

The probabilities of all the possible outcomes add up to $1$

Something must happen. If you list every possible outcome, one of them is certain to occur — and certainty has probability $1$.

Worked Example 1 — Finding a missing probability

A biased spinner has four colours. Find the missing probability.

Colour	Red	Blue	Green	White
Probability	$0.31$	$0.24$	$0.18$	$x$

①$0.31 + 0.24 + 0.18 + x = 1$

②$0.73 + x = 1$

③$x = 0.27$

Worked Example 2 — Two missing probabilities

A spinner has three sections. $P(\text{red}) = 0.4$, and green and yellow are equally likely. Find $P(\text{green})$.

①Let $P(\text{green}) = P(\text{yellow}) = x$.

②$0.4 + x + x = 1$

③$2x = 0.6 \Rightarrow x = 0.3$

$P(\text{green}) = 0.3$

Worked Example 3 — Probabilities in a ratio

A spinner has red, blue and green sections with probabilities in the ratio $3 : 4 : 5$. Find each probability.

①Total parts $= 3 + 4 + 5 = 12$.

②The total probability is $1$, so one part $= \dfrac{1}{12}$.

③$P(\text{red}) = \dfrac{3}{12} = \dfrac{1}{4}$

④$P(\text{blue}) = \dfrac{4}{12} = \dfrac{1}{3}$

⑤$P(\text{green}) = \dfrac{5}{12}$

Check: $\dfrac{3}{12} + \dfrac{4}{12} + \dfrac{5}{12} = 1$ ✓

2 The Complement Rule

Complement

$P(\text{not } A) = 1 - P(A)$ often written $P(A') = 1 - P(A)$

An event either happens or it does not — there is no third option. So the two probabilities must add to $1$.

The complement is often the shortcut. "At least one" questions are usually far easier as $1 - P(\text{none})$ than by adding up every case.

Worked Example 4 — Using the complement

The probability that a train arrives on time is $0.87$. Find the probability that it is late.

①$P(\text{late}) = 1 - 0.87 = 0.13$

Worked Example 5 — "At least one"

A fair coin is tossed three times. Find the probability of getting at least one head.

①"At least one head" is the opposite of "no heads at all", i.e. three tails.

②$P(\text{TTT}) = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$

③$P(\text{at least one head}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$

Listing all the cases with one, two or three heads would take far longer and risk missing some.

3 Mutually Exclusive Events

Two events are mutually exclusive if they cannot both happen at the same time. Rolling a $2$ and rolling a $5$ on one die are mutually exclusive; rolling an even number and rolling a number over $3$ are not (because $4$ and $6$ are both).

The addition rule

For mutually exclusive events: $P(A \text{ or } B) = P(A) + P(B)$

Only add when the events cannot overlap. If they can both happen, simply adding counts the overlap twice and gives an answer that is too big — sometimes over $1$, which is an instant giveaway.

Worked Example 6 — Adding mutually exclusive probabilities

A bag has $5$ red, $4$ blue and $6$ green counters. One is picked at random. Find $P(\text{red or green})$.

①A counter cannot be both red and green, so the events are mutually exclusive.

②Total $= 5 + 4 + 6 = 15$

③$P(\text{red or green}) = \dfrac{5}{15} + \dfrac{6}{15} = \dfrac{11}{15}$

Alternatively, $P(\text{not blue}) = 1 - \dfrac{4}{15} = \dfrac{11}{15}$ — the same answer via the complement.

Worked Example 7 — Events that overlap

A fair die is rolled. $A$ is "an even number" and $B$ is "a number greater than $3$". Are $A$ and $B$ mutually exclusive? Find $P(A \text{ or } B)$.

①$A = \{2, 4, 6\}$ and $B = \{4, 5, 6\}$.

②They share $4$ and $6$, so they are not mutually exclusive.

③Simply adding would give $\dfrac{3}{6} + \dfrac{3}{6} = 1$, which is wrong — rolling a $1$ satisfies neither.

④List the outcomes in $A$ or $B$: $\{2, 4, 5, 6\}$ — four outcomes.

⑤$P(A \text{ or } B) = \dfrac{4}{6} = \dfrac{2}{3}$

The general rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$: add them, then subtract the overlap once. Here $\tfrac{3}{6} + \tfrac{3}{6} - \tfrac{2}{6} = \tfrac{4}{6}$ ✓

4 Exhaustive Events

A set of events is exhaustive if it covers every possible outcome — at least one of them is bound to happen.

The key result

If a set of events is both mutually exclusive and exhaustive,
their probabilities add to exactly $1$

Set of events for one die roll	Mutually exclusive?	Exhaustive?
$\{1\}, \{2\}, \{3\}, \{4\}, \{5\}, \{6\}$	Yes	Yes
"even", "odd"	Yes	Yes
"less than 3", "more than 4"	Yes	No — $3$ and $4$ are missed
"even", "more than 3"	No — $4$ and $6$ are in both	No — $1$ and $3$ are missed
"less than 5", "more than 2"	No — $3$ and $4$ in both	Yes

Worked Example 8 — Testing a set of events

A card is drawn from a pack. Are the events "a heart", "a diamond" and "a black card" mutually exclusive and exhaustive?

①No card can be two of these at once, so they are mutually exclusive.

②Hearts and diamonds are all the red cards; "a black card" covers the rest.

③Every card falls into exactly one category, so they are exhaustive.

④Check: $\dfrac{13}{52} + \dfrac{13}{52} + \dfrac{26}{52} = 1$ ✓

5 Quick Reference

Total is 1

All possible outcomes have probabilities summing to $1$.

Complement

$P(\text{not } A) = 1 - P(A)$.

Mutually exclusive

Cannot both happen; no overlap.

Addition rule

Mutually exclusive: $P(A \text{ or } B) = P(A) + P(B)$.

With overlap

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

Exhaustive

Covers every possible outcome.

Both together

Mutually exclusive and exhaustive → probabilities total exactly $1$.

"At least one"

Use $1 - P(\text{none})$.

Ratios

Total the parts; each part is worth $\dfrac{1}{\text{total parts}}$.

6 Practice Questions

Question 1

A spinner has $P(\text{red}) = 0.15$, $P(\text{blue}) = 0.42$ and $P(\text{green}) = x$. Find $x$.

▶ Show solution

$0.15 + 0.42 + x = 1$

$x = 1 - 0.57 = 0.43$

Question 2

The probability that a student is left-handed is $\dfrac{2}{15}$. Find the probability that a student is not left-handed.

▶ Show solution

$P(\text{not left-handed}) = 1 - \dfrac{2}{15} = \dfrac{13}{15}$

Question 3

A bag has $6$ red, $8$ blue and $11$ yellow beads. Find (a) $P(\text{red or blue})$, (b) $P(\text{not yellow})$.

▶ Show solution

Total $= 6 + 8 + 11 = 25$

(a) $\dfrac{6}{25} + \dfrac{8}{25} = \dfrac{14}{25}$

(b) $1 - \dfrac{11}{25} = \dfrac{14}{25}$ — the same, as it must be.

Question 4

Are these pairs mutually exclusive? (a) Rolling a $3$ and rolling a $5$. (b) Drawing a king and drawing a heart. (c) A person being over $18$ and being under $12$.

▶ Show solution

(a) Yes — one roll cannot be both.

(b) No — the king of hearts is both.

(c) Yes — nobody can be both over $18$ and under $12$.

Question 5

A spinner has four sections with probabilities in the ratio $2 : 3 : 4 : 6$. Find the probability of each.

▶ Show solution

Total parts $= 2 + 3 + 4 + 6 = 15$.

Probabilities: $\dfrac{2}{15}$, $\dfrac{3}{15} = \dfrac{1}{5}$, $\dfrac{4}{15}$, $\dfrac{6}{15} = \dfrac{2}{5}$

Check: they add to $\dfrac{15}{15} = 1$ ✓

Question 6

A fair coin is tossed four times. Find the probability of getting at least one tail.

▶ Show solution

The opposite of "at least one tail" is "no tails", i.e. four heads.

$P(\text{HHHH}) = \left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}$

$P(\text{at least one tail}) = 1 - \dfrac{1}{16} = \dfrac{15}{16}$

Question 7

A die is rolled. $A$ = "a multiple of $3$" and $B$ = "an odd number". Find $P(A)$, $P(B)$ and $P(A \text{ or } B)$.

▶ Show solution

$A = \{3, 6\}$, so $P(A) = \dfrac{2}{6} = \dfrac{1}{3}$

$B = \{1, 3, 5\}$, so $P(B) = \dfrac{3}{6} = \dfrac{1}{2}$

They overlap at $3$, so they are not mutually exclusive.

$A$ or $B$ is $\{1, 3, 5, 6\}$, so $P(A \text{ or } B) = \dfrac{4}{6} = \dfrac{2}{3}$

Check with the formula: $\dfrac{2}{6} + \dfrac{3}{6} - \dfrac{1}{6} = \dfrac{4}{6}$ ✓

Question 8

Give an example of two events for a single roll of a die that are exhaustive but not mutually exclusive.

▶ Show solution

"Less than $5$" $= \{1,2,3,4\}$ and "more than $2$" $= \{3,4,5,6\}$.

Exhaustive: together they cover $1,2,3,4,5,6$ — every outcome.

Not mutually exclusive: $3$ and $4$ appear in both.

(Note their probabilities add to $\tfrac{4}{6} + \tfrac{4}{6} = \tfrac{8}{6} > 1$ — a clear sign of overlap.)

Question 9

A bag contains only red, blue and green counters. $P(\text{red}) = 0.28$ and there are twice as many blue as green. Find $P(\text{blue})$.

▶ Show solution

Let $P(\text{green}) = x$, so $P(\text{blue}) = 2x$.

$0.28 + 2x + x = 1$

$3x = 0.72 \Rightarrow x = 0.24$

$P(\text{blue}) = 2 \times 0.24 = 0.48$

Check: $0.28 + 0.48 + 0.24 = 1$ ✓

Question 10

A five-sided spinner has sections $1$ to $5$. $P(1) = 0.1$, $P(2) = 0.25$, $P(4) = 0.15$, and $P(3)$ is three times $P(5)$.

(a) Find $P(3)$ and $P(5)$. (b) Find $P(\text{an odd number})$. (c) Find $P(\text{not } 2)$. (d) The spinner is spun $400$ times. How many $3$s would you expect?

▶ Show solution

(a) Let $P(5) = x$, so $P(3) = 3x$.

$0.1 + 0.25 + 3x + 0.15 + x = 1$

$0.5 + 4x = 1 \Rightarrow 4x = 0.5 \Rightarrow x = 0.125$

$P(5) = 0.125$ and $P(3) = 0.375$

(b) Odd numbers are $1$, $3$, $5$ — mutually exclusive, so add:

$0.1 + 0.375 + 0.125 = 0.6$

(c) $1 - 0.25 = 0.75$

(d) $0.375 \times 400 = 150$ threes

Exhaustive & Mutually Exclusive Events (P4) · GCSE Maths Revision · Created with MathJax