Grade 9 — Probability: Compound Events

Full lesson notes

Everything covered in this lesson, in one place - useful for revision or printing.

\( P(A \text{ then } B) = P(A) \times P(B) \)

Lesson 1 organised outcomes with Venn diagrams. This lesson handles what happens when two or more things happen one after the other — a coin tossed twice, three times, or a die rolled twice.

You will work with:

Theoretical probability with bigger sample spaces (a pack of 52 cards)
Relative frequency and expected number
Two-way tables and tree diagrams for compound events
Trees where the branches are not equally likely

The big idea: multiply along a path, add across paths.

1. Revision

\( P(E) = \dfrac{\text{favourable outcomes}}{\text{possible outcomes}} \)

The outcomes must be equally likely for this to work.

A bag holds 3 red, 4 blue and 7 black pens — 14 pens altogether.
\(P(\text{red}) = \dfrac{3}{14}\), \(P(\text{blue}) = \dfrac{4}{14} = \dfrac{2}{7}\), \(P(\text{black}) = \dfrac{7}{14} = \dfrac{1}{2}\).

The denominator never changes — there are always 14 pens.

\( 0 \;\leq\; P(E) \;\leq\; 1 \)

Every probability sits on this line. If your answer is negative or bigger than 1, the arithmetic went wrong.

\( P(\text{certain}) = 1 \qquad P(\text{impossible}) = 0 \)

Roll one die:

\(P(\text{a natural number}) = \dfrac{6}{6} = 1\) — every face qualifies, so it is certain.
\(P(\text{greater than } 6) = \dfrac{0}{6} = 0\) — no face qualifies, so it is impossible.

\( 4 \times 13 = 52 \text{ cards} \)

A pack has four suits — clubs, spades, hearts, diamonds — with 13 cards in each.

\(P(\text{queen}) = \dfrac{4}{52} = \dfrac{1}{13}\) — there are four queens.
\(P(\text{queen of hearts}) = \dfrac{1}{52}\) — there is only one.

Read the question carefully: "a queen" and "the queen of hearts" are very different counts.

\( P(\text{spade or club}) = \dfrac{13}{52} + \dfrac{13}{52} = \dfrac{1}{2} \)

No card is both a spade and a club, so the two groups simply add together.

\(P(\text{a diamond, not } K, Q, J) = \dfrac{10}{52} = \dfrac{5}{26}\) — 13 diamonds less the three picture cards.

2. Relative Frequency

\( \text{relative frequency} = \dfrac{\text{times it happened}}{\text{number of trials}} \)

This is measured, not calculated. A die was actually thrown 50 times:

Number	1	2	3	4	5	6
Throws	5	7	8	10	13	7

The number 5 came up 13 times.

\( \dfrac{13}{50} = 0{,}26 \quad\text{but}\quad \dfrac{1}{6} \approx 0{,}17 \)

These do not match — and nothing is wrong.

50 throws is a small experiment. Chance alone pushes the result up and down.

Throw the die 5 000 times and the relative frequency creeps towards \(\dfrac{1}{6}\). More trials, closer agreement.

\( \text{not equally likely} \Rightarrow \text{use relative frequency} \)

The counting formula assumes every outcome is equally likely. A drawing pin, a bent coin, a real survey of people — none of those qualify.

For those you run the experiment and use what actually happened.

3. Expected Number

\( \text{expected number} = P(E) \times \text{number of trials} \)

Probability predicts how often something will happen over many trials.

A die is rolled 300 times. How many 5s?
\(300 \times \dfrac{1}{6} = 50\) fives.

\( 4000 \times \dfrac{1}{2} = 2000 \)

A die is rolled 4 000 times. How many odd numbers do you expect?
Half the faces are odd (1, 3, 5), so \(P(\text{odd}) = \dfrac{1}{2}\).

Expect about 2 000 odd numbers.

"Expect" is a prediction, not a promise — you will not get exactly 2 000.

4. Two-Way Tables

\( \text{simple event} + \text{simple event} = \text{compound event} \)

Tossing a coin once is a simple event. Tossing it twice is a compound event.

There are two ways to show every outcome: a two-way table or a tree diagram.

\( S = \{HH,\ HT,\ TH,\ TT\} \)

A coin tossed twice, as a grid:

	2nd: H	2nd: T
1st: H	HH	HT
1st: T	TH	TT

Four outcomes, each equally likely, so each has probability \(\dfrac{1}{4}\).

\( HT \neq TH \)

HT and TH are different outcomes. A head then a tail is not the same event as a tail then a head.

This is the single most common mistake — learners say there are 3 outcomes (two heads, two tails, one of each) and get every answer wrong.

5. Tree Diagrams

\( \text{each path} = \text{one outcome} \)

Follow one path from left to right. Each complete path is a single outcome.

\( \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} \)

The two tosses happen one after the other, so the fractions along a path multiply.

\(P(HH) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\) — the same answer the table gave.

\( \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \)

"A head and a tail in any order" is satisfied by two different paths: HT and TH.
They are separate outcomes, so their probabilities add.

\(P(\text{one of each}) = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}\)

\( \text{along} \rightarrow \times \qquad \text{across} \rightarrow + \)

Moving ALONG one path — the events happen in sequence, so multiply.

Collecting SEVERAL finished paths — they are alternatives, so add.

Every compound-event question is one of these two, or both.

\( P(\text{at least one head}) = \dfrac{3}{4} \)

Three of the four paths contain a head: HH, HT, TH.

Faster: the only path with no head is TT, so \(1 - \dfrac{1}{4} = \dfrac{3}{4}\).

6. Three Tosses

\( 2 \times 2 \times 2 = 8 \text{ outcomes} \)

Each extra toss doubles the paths: 2, then 4, then 8. Every path has probability \(\dfrac{1}{8}\).

\( P(HHH) = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8} \)

Multiply along the path, exactly as before — just three fractions instead of two.

\(P(TTH) = \dfrac{1}{8}\) too. Every single named outcome is one eighth.

\( \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} = \dfrac{3}{8} \)

"Two heads and a tail, in any order" happens on three paths: HHT, HTH and THH.

Find all the paths first, then add. Missing one is where the marks go.

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

The long way lists seven outcomes. The short way asks the opposite question:

No tails at all is only HHH, which is \(\dfrac{1}{8}\). Subtract from 1.

When a question says "at least one", reach for the complement first.

7. Unequal Branches

\( P(A) = \dfrac{1}{6} \qquad P(\text{not } A) = \dfrac{5}{6} \)

A die is rolled twice. Let \(A\) be getting a 4.
Now the two branches are not equally likely — one face out of six is a 4, and the other five are not.

These two must always add to 1.

\( \dfrac{1}{36} + \dfrac{5}{36} + \dfrac{5}{36} + \dfrac{25}{36} = 1 \)

The same rule applies — multiply along each path. The four end-probabilities must total 1, which is your check.

\( \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{36} \)

A 4 then not a 4: \(\dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{36}\)
Not a 4 both times: \(\dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{25}{36}\)

Notice the two middle paths give the same answer — but they are still two separate outcomes.

\( 1 - \dfrac{25}{36} = \dfrac{11}{36} \)

Method 1 (quicker): the only path with no 4 at all is \(\dfrac{25}{36}\). Subtract from 1.
Method 2: add the three paths that contain a 4: \(\dfrac{1}{36} + \dfrac{5}{36} + \dfrac{5}{36} = \dfrac{11}{36}\).

Two correct methods, one answer. If they disagree, one has an arithmetic slip.

\( P(HH) \neq \dfrac{1}{2} + \dfrac{1}{2} \)

1. Treating HT and TH as one outcome.

2. Adding along a path instead of multiplying — that gives 1, which would mean two heads are certain.

3. Leaving an answer above 1. Always sense-check against the 0-to-1 line.

\( \text{along} \rightarrow \times \qquad \text{across} \rightarrow + \)

Work out theoretical probability, including a 52-card pack
Use relative frequency, and say why it differs from theory
Predict an expected number over many trials
Build a two-way table and a tree diagram
Handle trees whose branches are not equally likely
Use the complement for "at least one"

Now try the quiz — it draws 15 questions from a bank of 30.