Mutually exclusive events
🎯 In this topic you will
- Identify mutually exclusive outcomes
- Use equally likely outcomes to calculate theoretical probabilities
- Use addition to calculate probabilities
- Apply the rule that the total probability of all possible mutually exclusive events is 1
🧠 Key Words
- equally likely
- mutually exclusive
- theoretical probability
Show Definitions
- equally likely: When two or more outcomes have the same chance of occurring.
- mutually exclusive: Events that cannot happen at the same time, such as rolling a 3 and a 5 on a single dice throw.
- theoretical probability: The probability of an event calculated using reasoning or mathematical models, assuming all outcomes are equally likely.
🔹 Mutually Exclusive Outcomes
When two outcomes cannot happen at the same time they are mutually exclusive. For example, if you get a $1$ when you throw a dice, you cannot also get a $3$.
🎲 Equally Likely Outcomes
When you roll a dice, each number is equally likely. $1, 2, 3, 4, 5$ and $6$ are equally likely outcomes.
⚽ Real-World Example
For the event of winning a football match, do you think that winning $1 - 0$ and winning $6 - 0$ are equally likely outcomes?
📊 Theoretical Probability
When the outcomes are equally likely, you do not need to do an experiment to find the probabilities. You can use the fact that outcomes are equally likely to calculate theoretical probabilities.
📊 Probabilities of Mutually Exclusive Outcomes
In Worked example 13.2 there are five mutually exclusive outcomes. They are the numbers $2$, $3$, $5$, $8$ and $9$. Here are the probabilities for each outcome.
| Outcome | $2$ | $3$ | $5$ | $8$ | $9$ |
|---|---|---|---|---|---|
| Probability | $10\%$ | $10\%$ | $40\%$ | $30\%$ | $10\%$ |
The sum of these probabilities is
$10\% + 10\% + 40\% + 30\% + 10\% = 100\% = 1$.
The probabilities add up to $1$ because exactly one of the numbers is certain to be chosen.
❓ EXERCISES
1. Chen throws a fair coin.
a. What are the possible outcomes?
b. Are the outcomes mutually exclusive?
c. Are the outcomes equally likely?
👀 Show answer
a. Outcomes: Heads and Tails (often written $H$ and $T$).
b. Yes. You cannot get Heads and Tails on the same single throw, so they are mutually exclusive.
c. Yes. For a fair coin, $P(H)=P(T)=\tfrac12$; the outcomes are equally likely.
2. A weather forecast says that the probability of rain tomorrow is $20\%$ and the probability of strong winds tomorrow is $30\%$.
a. Explain why rain and strong winds are not mutually exclusive outcomes.
b. Explain why rain and strong winds are not equally likely outcomes.
👀 Show answer
a. They can happen at the same time (it can be rainy and windy together), so they are not mutually exclusive.
b. Their probabilities are different: $20\% \ne 30\%$, so the outcomes are not equally likely.
3. Balsem throws an unbiased dice.
Work out the probability that she throws:
a. the number $4$
b. the number $3$
c. an odd number
d. an even number
e. the number $3$ or more
f. a number more than $6$
👀 Show answer
a.$P(4)=\tfrac{1}{6}$
b.$P(3)=\tfrac{1}{6}$
c. Odd numbers: $\{1,3,5\}$ so $P(\text{odd})=\tfrac{3}{6}=\tfrac12$
d. Even numbers: $\{2,4,6\}$ so $P(\text{even})=\tfrac{3}{6}=\tfrac12$
e. Numbers $\ge 3$: $\{3,4,5,6\}$ so $P(\ge 3)=\tfrac{4}{6}=\tfrac{2}{3}$
f. More than $6$: impossible on a standard die, so $P(>6)=0$
❓ EXERCISES
4. Each letter of the word MATHEMATICS is written on a separate card. The cards are placed face down. One card is chosen without looking. Work out the probability that the card is:
a. $H$
b. $A$
c. $S$
d. $M$
e. $B$
f. a letter in the word SCIENCE
👀 Show answer
a.$P(H)=\tfrac{1}{11}$
b.$P(A)=\tfrac{2}{11}$
c.$P(S)=\tfrac{1}{11}$
d.$P(M)=\tfrac{2}{11}$
e.$P(B)=0$ (no $B$ in the word)
f. Letters of SCIENCE are $\{S,C,I,E,N\}$. In MATHEMATICS we have $S,C,I,E$ once each, $N$ none ⇒ $4$ favourable cards. So $P=\tfrac{4}{11}$.
5. A coloured spinner has $12$ equal sections. When spun, the pointer has an equal likelihood of stopping in any of the sectors.
a. Work out the probability outcome of each of the four colours.
b. Show that the sum of the probabilities in part a is $1$.
👀 Show answer
(Use the actual counts from your spinner.)
b. The sum is $\dfrac{5}{12}+\dfrac{3}{12}+\dfrac{2}{12}+\dfrac{2}{12}=\dfrac{12}{12}=1$ (in general, the sector counts add to $12$, so probabilities add to $1$).
6. The letters of the word BANANA are printed on cards. The cards are placed face down. A card is chosen without looking.
a. Find the probability that the card is the letter:
i. $B$ ii. $A$ iii. $N$
👀 Show answer
i.$P(B)=\tfrac{1}{6}$
ii.$P(A)=\tfrac{3}{6}=\tfrac12$
iii.$P(N)=\tfrac{2}{6}=\tfrac13$
These exhaust all possibilities, so $\tfrac{1}{6}+\tfrac12+\tfrac13=1$. They add to $1$ because exactly one of the letters must be chosen.7. The names of six boys and four girls are put in a bag. One of the boys’ names is Blake. One of the girls’ names is Crystal. One name is chosen without looking.
a. Work out the probability that the name chosen is:
i. Blake ii. a boy’s name iii. a girl’s name iv. not Crystal v. a boy’s name or a girl’s name
b. Explain why the chosen name is $50\%$ more likely to be a boy’s than a girl’s name.
👀 Show answer
i.$P(\text{Blake})=\tfrac{1}{10}$
ii.$P(\text{boy})=\tfrac{6}{10}=\tfrac{3}{5}$
iii.$P(\text{girl})=\tfrac{4}{10}=\tfrac{2}{5}$
iv.$P(\text{not Crystal})=\tfrac{9}{10}$
v.$P(\text{boy or girl})=1$ (these are all the names in the bag)
b. The chance of a boy is $\tfrac{3}{5}$ and of a girl is $\tfrac{2}{5}$. The relative increase is $\dfrac{\tfrac{3}{5}-\tfrac{2}{5}}{\tfrac{2}{5}}=\dfrac{\tfrac{1}{5}}{\tfrac{2}{5}}=\tfrac12=50\%$, so a boy’s name is $50\%$ more likely.
8. Here is what Marcus says:
“My football team can win, draw or lose a match. These are the only three outcomes. Winning is one of these outcomes. The probability that my team wins is $\tfrac{1}{3}$.”
What is incorrect with Marcus’ argument?
👀 Show answer
❓ EXERCISES
9. An unbiased dice with $20$ faces has the numbers from $1$ to $20$. The dice is thrown.
a. Work out the probability of throwing:
i. the number $3$ ii. a multiple of $3$ iii. an odd number iv. a factor of $30$
b. Work out the probability of throwing:
i. the number $6$ ii. a number less than $6$ iii. a number more than $6$
c. Explain why the three probabilities in part b must add up to $1$.
d. i. Describe an outcome that has a probability of $75\%$. ii. Describe an outcome that has a probability of $15\%$.
👀 Show answer
a.
i.$P(3)=\tfrac{1}{20}$
ii. Multiples of $3$ in $1\!-\!20$ are $\{3,6,9,12,15,18\}$ ($6$ outcomes) ⇒ $P=\tfrac{6}{20}=\tfrac{3}{10}$
iii. Ten odds ⇒ $P=\tfrac{10}{20}=\tfrac{1}{2}$
iv. Factors of $30$ in the range: $\{1,2,3,5,6,10,15\}$ ($7$ outcomes) ⇒ $P=\tfrac{7}{20}$
b.
i.$P(6)=\tfrac{1}{20}$
ii. Numbers $<6$ are $\{1,2,3,4,5\}$ ⇒ $P=\tfrac{5}{20}=\tfrac{1}{4}$
iii. Numbers 6$ are $\{7,\dots,20\}$ ($14$ outcomes) ⇒ $P=\tfrac{14}{20}=\tfrac{7}{10}$
c. The events “$6$”, “less than $6$”, and “more than $6$” are mutually exclusive and cover all outcomes; exactly one must occur, so their probabilities sum to $1$.
d.i. Any event with $15$ favourable outcomes, e.g. “a number $\le 15$” has probability $\tfrac{15}{20}=\tfrac{3}{4}=75\%$.
ii. Any event with $3$ favourable outcomes, e.g. “a number less than $4$” has probability $\tfrac{3}{20}=15\%$.
10. Here are eight number cards: $2,3,5,7,11,13,17,19$. The cards are placed face down. A card is chosen without looking.
a. Work out the probability that the card chosen is:
i. an odd number ii. a prime number iii. a factor of $100$
b. Work out the probability that the card chosen is:
i. $5$ ii. $5$ or less iii. $5$ or more
c. Zara says: “One of the outcomes in part b must happen and so the probabilities must add up to $1$.” Explain why Zara is incorrect.
👀 Show answer
a.
i. Seven odds ⇒ $\tfrac{7}{8}$
ii. All cards are prime ⇒ $\tfrac{8}{8}=1$
iii. Factors of $100$ in the set are $2$ and $5$ ⇒ $\tfrac{2}{8}=\tfrac{1}{4}$
b.i.$\tfrac{1}{8}$ ii. cards $\le5$ are $\{2,3,5\}$ ⇒ $\tfrac{3}{8}$ iii. cards $\ge5$ are $\{5,7,11,13,17,19\}$ ⇒ $\tfrac{6}{8}=\tfrac{3}{4}$
c. The events in part b are not mutually exclusive (e.g., “$5$” is included in both “$5$ or less” and “$5$ or more”). Because of this overlap, adding their probabilities double-counts outcomes, so the total need not be $1$.
11. A computer produces a two-digit random number from $00$ to $99$. The outcome of every pair of digits is equally likely. Work out the probability that:
a. both the digits are $6$
b. exactly one digit is $6$
c. neither of the digits is $6$
👀 Show answer
a. Only $66$ ⇒ $\tfrac{1}{100}$
b. First $6$ and second not $6$: $9$ ways; or first not $6$ and second $6$: $9$ ways ⇒ $\tfrac{18}{100}=\tfrac{9}{50}=0.18$
c. Complement of “at least one $6$” ⇒ $(\tfrac{9}{10})^2=\tfrac{81}{100}=0.81$
🧠 Think like a Mathematician
12. Caleb has a set of cards. Each card has a number on it. The cards are placed face down and he takes a card without looking.
The probability that the chosen card is $3$ is $\tfrac{1}{3}$.
The probability that the chosen card is $4$ is $\tfrac{1}{4}$.
a. Find a possible list of Caleb’s cards.
b. What can you say about the number of cards in the set?
c. An-Mei has a different set of cards, where each card has a number on it. She places them face down and takes a card without looking.
In her set, the probability that the chosen card is $4$ is $\tfrac{1}{4}$ and the probability that the chosen card is $5$ is $\tfrac{1}{5}$.
What can you say about An-Mei’s set?
d. Could An-Mei have the same set as Caleb?
👀 Show answer
- a. A possible set is {3, 3, 4, 4, 4, x, y, z, ...}. For example, {3, 3, 4, 4, 4, 6, 7, 8, 9, 10, 11, 12}. This gives 1/3 chance for 3 (say 4 out of 12) and 1/4 chance for 4 (3 out of 12).
- b. The total number of cards must be a common multiple of 3 and 4, since probabilities are 1/3 and 1/4. So the total must be a multiple of 12.
- c. In An-Mei’s set, 1/4 of the cards are 4s and 1/5 are 5s. The total number of cards must be a common multiple of 4 and 5, i.e. a multiple of 20.
- d. No. Caleb’s set requires a multiple of 12 cards, An-Mei’s requires a multiple of 20 cards. There is no common total number of cards that satisfies both given probabilities.
🎯 Mutually Exclusive Events with Multiples
There are $25$ balls, numbered from $1$ to $25$, in a bag.
One ball is taken out at random. Here are some possible events:
- F: The number on the ball is a multiple of $5$
- S: The number on the ball is a multiple of $7$
- N: The number on the ball is a multiple of $9$
These events are mutually exclusive. This means only one of them can happen at one time.
The multiples of $5$ in the bag are $5,10,15,20,25$, so the probability of event F is $P(F)=\dfrac{5}{25}=\dfrac{1}{5}$. Similarly, the probability of S, $P(S)=\dfrac{3}{25}$ and the probability of N, $P(N)=\dfrac{2}{25}$.
The probability that F does not happen is $1-\dfrac{1}{5}=\dfrac{4}{5}$. The probability that S does not happen is $1-\dfrac{3}{25}=\dfrac{22}{25}$.
What is the probability that F or S happens? This means you get $5,10,15,20,25,7,14$ or $21$. There are $8$ numbers, so the probability of F or S is $P(F\ \text{or}\ S)=\dfrac{8}{25}$. There is an easier way to work this out: just add the probabilities of the separate events.
- $P(F\ \text{or}\ S)=P(F)+P(S)=\dfrac{5}{25}+\dfrac{3}{25}=\dfrac{8}{25}$
- $P(S\ \text{or}\ N)=P(S)+P(N)=\dfrac{3}{25}+\dfrac{2}{25}=\dfrac{5}{25}=\dfrac{1}{5}$
- $P(F\ \text{or}\ S\ \text{or}\ N)=\dfrac{5}{25}+\dfrac{3}{25}+\dfrac{2}{25}=\dfrac{10}{25}=\dfrac{2}{5}$
The probability that none of F or S or N happens is $1-\dfrac{2}{5}=\dfrac{3}{5}$.
$P(F)$ means the probability of event F.
Addition works here because the events are mutually exclusive.
❓ EXERCISES
1. The probability a football team will win its next match is $60\%$.
The probability it will draw is $15\%$.
Work out the probability it will lose.
👀 Show answer
2. You roll a fair dice.
Work out the probability of rolling
a. a $3$
b. an even number
c. a $3$ or an even number.
👀 Show answer
a.$P(3)=\tfrac{1}{6}$
b. Even numbers $\{2,4,6\}$ ⇒ $P(\text{even})=\tfrac{3}{6}=\tfrac12$
c.$\{3\}\cup\{2,4,6\}=\{2,3,4,6\}$ ⇒ $P= \tfrac{4}{6}= \tfrac{2}{3}$
3. Here are $10$ numbered cards.
A card is chosen at random. Find the probability that the number on the card is
a. $2$
b. $5$
c. $2$ or $5$
d. neither $2$ nor $5$.
👀 Show answer
a.$P(2)=\tfrac{4}{10}=\tfrac{2}{5}$
b.$P(5)=\tfrac{3}{10}$
c.$P(2\text{ or }5)=\tfrac{4+3}{10}=\tfrac{7}{10}$
d. Remaining cards $=3$ ($3,3,4$) ⇒ $P=\tfrac{3}{10}$
4. A bag contains a large number of coloured balls. The balls are yellow, green, brown, blue and pink.
$P(\text{yellow})=0.1$ $P(\text{green})=0.25$ $P(\text{brown})=0.35$ $P(\text{blue})=0.05$
A ball is taken out of the bag at random.
Work out the probability that the ball is
a. green or blue
b. brown or yellow
c. yellow, green or brown
d. pink
👀 Show answer
a.$0.25+0.05=0.30$
b.$0.35+0.10=0.45$
c.$0.10+0.25+0.35=0.70$
d.$0.25$
❓ EXERCISES
5. You roll $2$ dice and find the total.
The probability of a total of $2$ is $0.028$.
The probability of a total of $3$ is $0.056$.
Work out the probability that the total is
a. $2$ or $3$
b. more than $3$.
👀 Show answer
a.$P(2\ \text{or}\ 3)=0.028+0.056=0.084$
b.$P(>3)=1-(0.028+0.056)=1-0.084=0.916$
6. The temperature each day at midday can be low, average, high or very high.
The forecast for Monday is $P(\text{low})=0.15$, $P(\text{average})=0.55$, $P(\text{high})=0.25$.
Work out the probability that the temperature on Monday is
a. not low
b. low or average
c. very high.
👀 Show answer
a.$1-0.15=0.85$
b.$0.15+0.55=0.70$
c.$1-(0.15+0.55+0.25)=0.05$
7. There are $50$ people in a room. There are $7$ girls, $13$ boys and $19$ women. The rest of the people are men.
One person is chosen at random. Work out the probability that the person is
a. a child
b. a female
c. an adult
d. a male.
👀 Show answer
a.$\tfrac{20}{50}=\tfrac{2}{5}=0.4$
b.$\tfrac{7+19}{50}=\tfrac{26}{50}=\tfrac{13}{25}=0.52$
c.$\tfrac{11+19}{50}=\tfrac{30}{50}=\tfrac{3}{5}=0.6$
d.$\tfrac{13+11}{50}=\tfrac{24}{50}=\tfrac{12}{25}=0.48$
8. The letters of the word MUTUALLY are written on separate cards. One card is chosen at random. Work out the probability that the letter on the card is
a. $M$
b. $U$
c. $L$
d. $M, U$ or $L$
e. not $M, U$ or $L$.
👀 Show answer
a.$\tfrac{1}{8}$
b.$\tfrac{2}{8}=\tfrac{1}{4}$
c.$\tfrac{2}{8}=\tfrac{1}{4}$
d.$\tfrac{1+2+2}{8}=\tfrac{5}{8}$
e. Remaining letters $T,A,Y$ ⇒ $\tfrac{3}{8}$
9. There are red, white, green and black counters in a box. A counter is taken out at random.
The probability the counter is red is $0.55$.
The other three colours are all equally likely.
Find the probability the counter is
a. not red
b. red or white.
👀 Show answer
a.$0.45$
b.$0.55+0.15=0.70$
10. A spinner has three sectors labelled $A$, $B$ and $C$.
The probability of landing on $A$ is twice the probability of landing on $B$.
The probability of landing on $B$ is twice the probability of landing on $C$.
Work out the probability of landing on each letter.
👀 Show answer
$4x+2x+x=7x=1 \Rightarrow x=\tfrac{1}{7}$.
So $P(A)=\tfrac{4}{7}$, $P(B)=\tfrac{2}{7}$, $P(C)=\tfrac{1}{7}$.
❓ EXERCISES
11. There are counters of four different colours in a bag. A counter is taken out at random.
This table shows the probabilities of different colours.
| Colour | Probability |
|---|---|
| gold | $0.05$ |
| silver | $0.15$ |
| bronze | $0.25$ |
| white | $0.55$ |
Find the probability that the counter is
a. gold or silver
b. not gold
c. silver or bronze.
👀 Show answer
a. $0.05+0.15=0.20$
b. $1-0.05=0.95$
c. $0.15+0.25=0.40$
🧠 Reasoning Tip
‘Between $1$ and $100$ inclusive’ means $1$ and $100$ are included in the possible numbers.
12. A calculator generates a random number between $1$ and $100$ inclusive.
Find the probability the number is
a. a multiple of $10$
b. a multiple of $11$
c. a multiple of $10$ or $11$
d. not a multiple of $10$ or $11$.
👀 Show answer
a. Multiples of $10$: $10,20,\dots,100$ ⇒ $10$ numbers ⇒ $P=\tfrac{10}{100}=0.10$
b. Multiples of $11$: $11,22,\dots,99$ ⇒ $9$ numbers ⇒ $P=\tfrac{9}{100}=0.09$
c. Use inclusion–exclusion:
$10+9-1=18$ numbers (LCM $110$ beyond range).
$P=\tfrac{18}{100}=0.18$
d. $1-0.18=0.82$
🧠 Think like a Mathematician
13. You want to put black, white and yellow counters in a bag. One counter will be chosen at random.
You want $P(\text{black})=\tfrac{1}{4}$ and $P(\text{white})=\tfrac{2}{3}$.
a. Work out $P(\text{yellow})$.
b. What is a suitable number of counters of each colour to put in the bag? Give a reason for your answer.
👀 Show answer
- a. Since probabilities must sum to $1$, $P(\text{yellow})=1-\tfrac{1}{4}-\tfrac{2}{3}=\tfrac{1}{12}$.
- b. A common denominator of $12$ works. $P(\text{black})=\tfrac{3}{12}$, $P(\text{white})=\tfrac{8}{12}$, $P(\text{yellow})=\tfrac{1}{12}$. So a suitable distribution is 3 black counters, 8 white counters and 1 yellow counter (out of 12 total). This satisfies the given probabilities exactly.