Compound probability - Year 9 Mathematics

What you will learn

list outcomes of two-step experiments using tree diagrams and tables,
distinguish between experiments with replacement and without replacement,
calculate probabilities of compound events using the “and” (multiplication) and “or” (addition) rules,
connect experimental (relative frequency) results with theoretical probability,
use simulations to estimate probabilities when calculation is difficult.

Worked example 0 Real-world example: quality control

A box contains $8$ good batteries and $2$ faulty ones. Two batteries are selected at random without replacement. What is the probability that both are good?

$P(\text{1st good}) = \dfrac{8}{10}$ .
After removing one good battery, $7$ good remain out of $9$ total: $P(\text{2nd good} \mid \text{1st good}) = \dfrac{7}{9}$ .
$P(\text{both good}) = \dfrac{8}{10} \times \dfrac{7}{9} = \dfrac{56}{90} = \dfrac{28}{45} \approx 0.622$ .

Key idea: without replacement, the second probability changes because the total and the count of favourable outcomes both decrease.

1. Two-step experiments and tree diagrams

A compound event involves two or more stages. A tree diagram shows every possible path from start to finish, with probabilities written on the branches.

Tree diagram: drawing 2 marbles without replacement from a bag containing 3 red and 2 blue marbles.

To find the probability of any single path, multiply the probabilities along the branches. The four outcomes sum to $1$ :

$\dfrac{6}{20} + \dfrac{6}{20} + \dfrac{6}{20} + \dfrac{2}{20} = \dfrac{20}{20} = 1.$

2. With replacement vs without replacement

Feature	With replacement	Without replacement
The item is returned before the next draw	Yes	No
Total stays the same	Yes	No — decreases by 1 each draw
Probabilities change between draws	No	Yes
Events are	Independent	Dependent

Worked example 1 With replacement

A bag contains $4$ red and $6$ blue marbles. Two marbles are drawn with replacement. Find $P(\text{both red})$ .

$P(\text{1st red}) = \dfrac{4}{10} = \dfrac{2}{5}$ .
The marble is replaced, so $P(\text{2nd red}) = \dfrac{4}{10} = \dfrac{2}{5}$ .
$P(\text{both red}) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}$ .

Worked example 2 Without replacement

Same bag ( $4$ red, $6$ blue). Two marbles drawn without replacement. Find $P(\text{both red})$ .

$P(\text{1st red}) = \dfrac{4}{10}$ .
Now $3$ red remain out of $9$ : $P(\text{2nd red} \mid \text{1st red}) = \dfrac{3}{9} = \dfrac{1}{3}$ .
$P(\text{both red}) = \dfrac{4}{10} \times \dfrac{1}{3} = \dfrac{4}{30} = \dfrac{2}{15}$ .

Note: $\dfrac{2}{15} < \dfrac{4}{25}$ — removing a red marble on the first draw makes a second red less likely.

3. “And” vs “or” probabilities

Probability rules for compound events

And rule (multiplication)

$P(A \text{ and } B) = P(A) \times P(B \mid A)$ If $A$ and $B$ are independent: $P(A \text{ and } B) = P(A) \times P(B)$ .

Or rule (addition)

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ If $A$ and $B$ are mutually exclusive: $P(A \text{ or } B) = P(A) + P(B)$ .

Worked example 3 Using the 'or' rule

A standard deck of $52$ cards. One card is drawn. Find $P(\text{heart or king})$ .

$P(\text{heart}) = \dfrac{13}{52}$ , $P(\text{king}) = \dfrac{4}{52}$ .
$P(\text{heart and king}) = \dfrac{1}{52}$ (the king of hearts).
$P(\text{heart or king}) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$ .

4. Relative frequency and simulations

Theoretical probability is calculated from known, equally likely outcomes. Experimental probability (or relative frequency) is calculated from observed data:

Relative frequency

$P(\text{event}) \approx \dfrac{\text{number of times event occurred}}{\text{total number of trials}}$

As the number of trials increases, relative frequency tends to approach theoretical probability — this is the law of large numbers.

A simulation models a random experiment using a tool (coin flips, random number generators, spreadsheets) when theoretical calculation is difficult or impossible.

Worked example 4 Simulation design

Estimate the probability that a family with $3$ children has exactly $2$ girls.

Model each child as a coin flip: heads $=$ girl, tails $=$ boy.
Flip $3$ coins and record whether exactly $2$ are heads. This is one trial.
Repeat for $100$ trials. Suppose $38$ trials gave exactly $2$ heads.
Relative frequency $= \dfrac{38}{100} = 0.38$ .
Theoretical probability $= \dbinom{3}{2} \times \left(\dfrac{1}{2}\right)^3 = 3 \times \dfrac{1}{8} = \dfrac{3}{8} = 0.375$ .

The simulation result ( $0.38$ ) is close to the theoretical value ( $0.375$ ).

Practice

Fluency

Tier 1: basic skills

A coin is tossed twice. List all outcomes using a tree diagram.
Two dice are rolled. How many outcomes are in the sample space?
A bag has $5$ red and $3$ green marbles. One marble is drawn and replaced, then another is drawn. Find $P(\text{both red})$ .
Repeat Q3 but without replacement.
A spinner has sections labelled $1, 2, 3, 4$ (equally likely). It is spun twice. Find $P(\text{both even})$ .
From a standard deck of $52$ cards, one card is drawn. Find $P(\text{spade or ace})$ .
A coin is tossed $200$ times and lands heads $114$ times. Calculate the relative frequency of heads.
Are the events “rolling a $6$ ” and “rolling an odd number” on a single die mutually exclusive? Explain.
A tree diagram has two branches at the first stage ( $A$ and $B$ ) and two at the second stage ( $C$ and $D$ from each). How many outcomes are there in total?
In $50$ trials of a simulation, an event occurred $18$ times. Estimate the probability of the event.

Reasoning

Tier 2: mixed practice

A bag contains $6$ red, $4$ blue, and $2$ green marbles. Two marbles are drawn without replacement. Draw a full tree diagram and find $P(\text{both blue})$ .
Two cards are drawn without replacement from a standard deck. Find $P(\text{both aces})$ .
A box has $3$ defective items out of $20$ . Two items are selected at random without replacement. Find the probability that (a) both are defective, (b) neither is defective, (c) exactly one is defective.
Events $A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.5$ . Find $P(A \text{ and } B)$ and $P(A \text{ or } B)$ .
A student rolls a die and flips a coin. Find the probability of getting an even number and heads.
In a class of $30$ , there are $12$ students who play sport and $8$ who play music. Of these, $3$ play both. A student is chosen at random. Find $P(\text{sport or music})$ .
Explain the difference between independent events and mutually exclusive events, using an example of each.
Design a simulation to estimate the probability that when $4$ coins are tossed, at least $3$ are tails. Describe the steps clearly.

Reasoning

Tier 3: explain and apply

A medical test has a $95\%$ chance of correctly detecting a disease (sensitivity) and a $90\%$ chance of correctly identifying a healthy person (specificity). If $2\%$ of the population has the disease, draw a tree diagram and find the probability that a person who tests positive actually has the disease.
Two events satisfy $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \text{ or } B) = 0.7$ . Determine whether $A$ and $B$ are independent. Justify.
Three marbles are drawn without replacement from a bag of $5$ red and $4$ white. Find the probability of drawing at least one red marble.
A game costs $2 to play. You roll two dice: if the sum is $7$ you win $10, otherwise you win nothing. Find the expected profit per game and decide whether the game is fair.
Explain why $P(A \text{ or } B) \leq P(A) + P(B)$ , and state when equality holds.

Challenge

Reasoning

Harder reasoning

Five cards numbered $1$ to $5$ are placed face down. Two cards are selected at random without replacement. Find the probability that the sum of the two cards is even.
A bag contains $n$ red and $3$ blue marbles. Two marbles are drawn without replacement. If $P(\text{both red}) = \dfrac{1}{2}$ , find $n$ .
In a best-of-three game series, Team A has a $0.6$ probability of winning each game (games are independent). Draw a tree diagram and find the probability that Team A wins the series.
Three students independently attempt a problem. Their probabilities of solving it are $0.7$ , $0.8$ , and $0.9$ respectively. Find the probability that at least one student solves the problem.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Tier 1

Outcomes: HH, HT, TH, TT (4 outcomes).
$6 \times 6 = 36$ outcomes.
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{5}{8} = \dfrac{25}{64}$ .
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
Even numbers are $2$ and $4$ , so $P(\text{even}) = \dfrac{2}{4} = \dfrac{1}{2}$ . $P(\text{both even}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ .
$P(\text{spade}) = \dfrac{13}{52}$ , $P(\text{ace}) = \dfrac{4}{52}$ , $P(\text{spade and ace}) = \dfrac{1}{52}$ . $P(\text{spade or ace}) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$ .
Relative frequency $= \dfrac{114}{200} = 0.57$ .
Yes, they are mutually exclusive. A single die cannot show $6$ (even) and an odd number at the same time — the events have no outcomes in common.
$2 \times 2 = 4$ outcomes.
$P \approx \dfrac{18}{50} = 0.36$ .

Tier 2

Total $= 12$ . $P(\text{1st blue}) = \dfrac{4}{12} = \dfrac{1}{3}$ . $P(\text{2nd blue} \mid \text{1st blue}) = \dfrac{3}{11}$ . $P(\text{both blue}) = \dfrac{1}{3} \times \dfrac{3}{11} = \dfrac{3}{33} = \dfrac{1}{11}$ .
$P(\text{both aces}) = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{12}{2652} = \dfrac{1}{221}$ .
(a) $P(\text{both defective}) = \dfrac{3}{20} \times \dfrac{2}{19} = \dfrac{6}{380} = \dfrac{3}{190}$ . (b) $P(\text{neither defective}) = \dfrac{17}{20} \times \dfrac{16}{19} = \dfrac{272}{380} = \dfrac{68}{95}$ . (c) $P(\text{exactly one}) = 1 - \dfrac{3}{190} - \dfrac{68}{95} = 1 - \dfrac{3}{190} - \dfrac{136}{190} = \dfrac{51}{190} = \dfrac{51}{190}$ .
$P(A \text{ and } B) = 0.3 \times 0.5 = 0.15$ . $P(A \text{ or } B) = 0.3 + 0.5 - 0.15 = 0.65$ .
$P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$ . $P(\text{heads}) = \dfrac{1}{2}$ . Events are independent, so $P(\text{even and heads}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ .
$P(\text{sport or music}) = \dfrac{12 + 8 - 3}{30} = \dfrac{17}{30}$ .
Independent events: the occurrence of one does not affect the probability of the other. Example: rolling a die and flipping a coin — the die result does not change the coin probability. Mutually exclusive events: the events cannot both occur at the same time. Example: rolling a $3$ and rolling a $5$ on a single die. Note: mutually exclusive events with non-zero probabilities are never independent (if one occurs, the probability of the other becomes $0$ ).
Simulation steps: (i) Assign heads $=$ tails outcome for a coin. (ii) Flip $4$ coins and record the number of tails. (iii) If $3$ or $4$ tails, record a success. (iv) Repeat for $100$ or more trials. (v) Estimate $P(\text{at least 3 tails}) = \dfrac{\text{number of successes}}{\text{total trials}}$ . Theoretical value: $P = \dbinom{4}{3}\left(\dfrac{1}{2}\right)^4 + \dbinom{4}{4}\left(\dfrac{1}{2}\right)^4 = \dfrac{4+1}{16} = \dfrac{5}{16} = 0.3125$ .

Tier 3

Let $D =$ has disease, $T^+ =$ tests positive. $P(D) = 0.02$ , $P(T^+ \mid D) = 0.95$ , $P(T^+ \mid \text{no } D) = 0.10$ . By tree diagram: $P(T^+) = 0.02 \times 0.95 + 0.98 \times 0.10 = 0.019 + 0.098 = 0.117$ . $P(D \mid T^+) = \dfrac{0.019}{0.117} \approx 0.162$ . Only about $16\%$ of positive results are true positives — the low disease rate means most positives are false alarms.
$P(A \text{ and } B) = P(A) + P(B) - P(A \text{ or } B) = 0.4 + 0.5 - 0.7 = 0.2$ . If independent: $P(A) \times P(B) = 0.4 \times 0.5 = 0.2$ . Since $P(A \text{ and } B) = P(A) \times P(B)$ , yes, $A$ and $B$ are independent.
It is easier to find $P(\text{no red}) = P(\text{all white}) = \dfrac{4}{9} \times \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{24}{504} = \dfrac{1}{21}$ . So $P(\text{at least one red}) = 1 - \dfrac{1}{21} = \dfrac{20}{21}$ .
$P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$ . Expected winnings $= \dfrac{1}{6} \times 10 + \dfrac{5}{6} \times 0 = \dfrac{10}{6} \approx 1.67$ dollars. Expected profit $= 1.67 - 2 = -0.33$ dollars. The game is not fair — on average, the player loses about $33$ cents per game.
$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ . Since $P(A \text{ and } B) \geq 0$ , we have $P(A \text{ or } B) \leq P(A) + P(B)$ . Equality holds when $P(A \text{ and } B) = 0$ , i.e. when $A$ and $B$ are mutually exclusive.

Challenge

Cards $1$ – $5$ : odd numbers are $1, 3, 5$ (three), even numbers are $2, 4$ (two). For an even sum, both cards must be the same parity. $P(\text{both odd}) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20}$ . $P(\text{both even}) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20}$ . $P(\text{even sum}) = \dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}$ .
$P(\text{both red}) = \dfrac{n}{n+3} \times \dfrac{n-1}{n+2} = \dfrac{1}{2}$ . So $2n(n-1) = (n+3)(n+2)$ , giving $2n^2 - 2n = n^2 + 5n + 6$ , i.e. $n^2 - 7n - 6 = 0$ . Using the quadratic formula: $n = \dfrac{7 + \sqrt{49 + 24}}{2} = \dfrac{7 + \sqrt{73}}{2} \approx \dfrac{7 + 8.544}{2} \approx 7.77$ . Since $n$ must be a positive integer, we check $n = 8$ : $\dfrac{8}{11} \times \dfrac{7}{10} = \dfrac{56}{110} = \dfrac{28}{55} \neq \dfrac{1}{2}$ . Check $n = 6$ : $\dfrac{6}{9} \times \dfrac{5}{8} = \dfrac{30}{72} = \dfrac{5}{12} \neq \dfrac{1}{2}$ . Since no integer solution exists, the equation $n^2 - 7n - 6 = 0$ has no positive integer root. Revisiting: if we allow $P(\text{both red}) = \dfrac{1}{2}$ to be approximate, $n = 8$ gives $\dfrac{28}{55} \approx 0.509$ , which is closest. However, for an exact solution: no integer value of $n$ works — this demonstrates that not every target probability is achievable with whole numbers of marbles.
Team A wins in 2 games: $P = 0.6 \times 0.6 = 0.36$ . Team A loses game 1, wins games 2 and 3: $P = 0.4 \times 0.6 \times 0.6 = 0.144$ . Team A wins game 1, loses game 2, wins game 3: $P = 0.6 \times 0.4 \times 0.6 = 0.144$ . Total: $P(\text{A wins series}) = 0.36 + 0.144 + 0.144 = 0.648$ .
$P(\text{none solve}) = (1 - 0.7)(1 - 0.8)(1 - 0.9) = 0.3 \times 0.2 \times 0.1 = 0.006$ . $P(\text{at least one solves}) = 1 - 0.006 = 0.994$ .

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