Lesson 17 ~30 min Unit 4 · Probability +85 XP

Theoretical Probability

Calculate exact probabilities without any experiment — using equally likely outcomes, the P = f/n formula, and the powerful complementary events rule.

Today's hook: If you know a coin is perfectly fair, you don't need to flip it 1000 times to know P(heads) = 1/2. Theoretical probability lets you calculate exactly — IF outcomes are equally likely. No experiment needed.

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Think First

warm-up

Before you read on — quickly: A bag contains 3 red balls and 2 blue balls. Without doing any experiment, what do you think the probability of picking a red ball is? How did you work that out?

Record your answer in your workbook.

The Big Idea

+5 XP

Theoretical probability is calculated mathematically, without any experiment, by assuming all outcomes are equally likely. The formula is: P(event) = number of favourable outcomes ÷ total number of outcomes.

A favourable outcome is any outcome that matches the event you want. The total outcomes is the complete count of all possible results in the sample space. The key assumption is that every outcome is equally likely — a fair die, a fair coin, drawing from a well-mixed bag.

P(event) = favourable outcomes ÷ total outcomes

Count carefully

List ALL outcomes first, then count the ones that match your event.

Check equally likely

The formula only works if every outcome has the same chance of happening.

Result is 0 to 1

P = 0 means impossible; P = 1 means certain. Any answer outside this range is wrong.

What You'll Master

objectives

Know

The formula P(event) = favourable ÷ total
P = 0 means impossible; P = 1 means certain
The complementary rule: P(A') = 1 − P(A)

Understand

Why outcomes must be equally likely for this formula to apply
Why a loaded die or biased spinner breaks the rule
When to use the complement instead of direct calculation

Can Do

List the sample space for a simple experiment
Calculate theoretical probability as a fraction, decimal or percentage
Use complementary events to find probability efficiently

Words You Need

vocabulary

Theoretical probabilityProbability calculated mathematically, assuming equally likely outcomes. No experiment needed.

Equally likelyEvery outcome has the same chance of occurring (e.g. fair die, fair coin).

Favourable outcomesThe outcomes that match the event you want to find the probability of.

Sample spaceThe complete set of ALL possible outcomes of an experiment.

Complementary eventsTwo events that together cover all outcomes. P(A') = 1 − P(A).

Impossible / CertainImpossible: P = 0 (can never happen). Certain: P = 1 (must always happen).

Spot the Trap

heads-up

Wrong: "There are 2 outcomes (win or lose) so P(win) = 1/2." Just because there are two outcomes does NOT mean they're equally likely.

Right: First verify equally likely. Winning a lottery and losing are not equally likely — you need experimental data or more information.

Wrong: Using P = f/n formula for a loaded die or biased spinner, where outcomes are NOT equally likely.

Right: Always check: is the die/coin/spinner fair? If biased, use experimental probability from real data instead.

Listing Outcomes and Finding Favourables

+5 XP

The first step is always to list the complete sample space — every possible outcome. Then identify which outcomes are favourable (match your event). Finally, apply P = f/n.

A standard die is rolled. Sample space: {1, 2, 3, 4, 5, 6} — 6 outcomes total. Event: rolling a prime. Primes on a die: {2, 3, 5} — 3 favourable outcomes. So P(prime) = 3/6 = 1/2. Remember: 1 is NOT prime.

P(prime) = 3 ÷ 6 = 1/2

List first

Always write out the sample space before counting. Missing an outcome leads to wrong answers.

Simplify the fraction

3/6 = 1/2. Always simplify unless asked to leave it unsimplified.

Know your primes

Primes to 20: 2, 3, 5, 7, 11, 13, 17, 19. Remember: 1 is NOT prime.

Computing Theoretical Probability

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Apply P = f/n where f = number of favourable outcomes and n = total outcomes. Express your answer as a fraction in lowest terms, a decimal, or a percentage.

A bag contains 4 red, 3 blue, and 5 green marbles. Total = 12. P(blue) = 3/12 = 1/4 = 0.25. P(not green) = (4+3)/12 = 7/12. You can also find P(not green) as 1 − P(green) = 1 − 5/12 = 7/12. Both methods work!

P = favourable ÷ total Simplify fractions

Add for total

Total = 4 + 3 + 5 = 12. Always add them explicitly — don't guess.

"Not" means others

P(not green) = red + blue = 7. Or use complement: 1 − P(green) = 7/12.

Probabilities sum to 1

P(red) + P(blue) + P(green) = 4/12 + 3/12 + 5/12 = 1. Check your work!

Complementary Events

+5 XP

The complement of event A (written A') is all outcomes where A does NOT happen. The rule is: P(A') = 1 − P(A). This is often the fastest approach for "at least one" problems.

Two coins are flipped. P(at least one head) — Complement method: P(no heads) = P(TT) = 1/4. So P(at least one head) = 1 − 1/4 = 3/4. Compare to direct method: list {HH, HT, TH, TT}; favourable = {HH, HT, TH} = 3 out of 4. Same answer!

P(A') = 1 − P(A) Always sums to 1

Use for "at least"

P(at least one) = 1 − P(none). Much easier than listing all favourable outcomes.

Complement notation

A' (A-prime) means "not A". The complement covers everything A does not.

Always sums to 1

P(A) + P(A') = 1. This is always true. Use it to verify your calculations.

Watch Me Solve It · 3 examples

Watch Me Solve It · P(prime) on a die

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PROBLEM

A standard die (faces 1–6) is rolled once. Find P(rolling a prime number).

1

List the sample space

{1, 2, 3, 4, 5, 6} — 6 outcomes total

A fair die: all outcomes equally likely. Check condition met.
2

Identify favourable outcomes

Primes: {2, 3, 5} — 3 favourable. (1 is not prime; 4 and 6 are composite)
3

Apply the formula

P(prime) = 3 ÷ 6 = 1/2 = 0.5 = 50%

A prime appears on exactly half the faces of a standard die.

AnswerP(prime) = 3/6 = 1/2 = 0.5

Watch Me Solve It · P(not green) from a bag

+15 XP per step

PROBLEM

A bag has 4 red, 3 blue and 5 green marbles. One marble is drawn at random. Find P(not green).

1

Find the total

Total = 4 + 3 + 5 = 12 marbles

Well-mixed bag: all marbles equally likely to be drawn.
2

Count favourables (not green)

Not green = red + blue = 4 + 3 = 7
3

Apply formula and check with complement

P(not green) = 7/12 Check: 1 − P(green) = 1 − 5/12 = 7/12 ✓

Both methods agree. 7/12 ≈ 0.583 = 58.3%.

AnswerP(not green) = 7/12 ≈ 0.583

Watch Me Solve It · Complement for "at least one"

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PROBLEM

Two fair coins are flipped. Find P(at least one head) using the complementary rule.

1

Identify the complement

Complement of "at least one head" = "no heads" = TT only

Finding P(no heads) is much easier than listing all "at least one head" outcomes.
2

List sample space and find P(complement)

Sample space: {HH, HT, TH, TT} — 4 outcomes. P(TT) = 1/4
3

Apply complement rule

P(at least one head) = 1 − 1/4 = 3/4

Verify directly: {HH, HT, TH} = 3 favourable out of 4. Correct!

AnswerP(at least one head) = 1 − 1/4 = 3/4 = 0.75

Common Pitfalls

heads-up

Not listing the complete sample space

For a die, if you only list the even numbers {2, 4, 6} when finding P(even), you'll get 3/3 = 1 instead of 3/6 = 1/2. The denominator must be ALL outcomes, not just the favourables.

Fix: Always write the full sample space first. The denominator is the total count of all outcomes.

Assuming equally likely when they're not

"It's heads or tails so P(heads) = 1/2." True for a fair coin — but if the coin is bent or weighted, this assumption fails. Never blindly use P = f/n without checking fairness.

Fix: Ask "is this a fair/unbiased situation?" If yes, use P = f/n. If no, you need experimental probability.

Forgetting "at least" → use complement

P(at least one tail in 3 flips) has 7 favourable outcomes. It's much faster to find P(no tails) = (1/2)³ = 1/8, then P(at least one tail) = 1 − 1/8 = 7/8.

Fix: When you see "at least", automatically think complement: 1 − P(none).

Copy Into Your Books

Theoretical Probability

P(event) = favourable ÷ total
Requires: equally likely outcomes
0 ≤ P ≤ 1 always
P = 0: impossible; P = 1: certain

Complementary Events

P(A') = 1 − P(A)
A' = "not A"
P(A) + P(A') = 1 always
Best for: "at least one" problems

Steps to Find P

1. List full sample space
2. Count favourable outcomes
3. Apply P = f/n
4. Simplify fraction

Check: All Probs = 1

Sum of all event probs = 1
Use to find unknown probabilities
P(A) + P(A') = 1

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Theoretical Probability

4 problems

Four drill problems. Work each, then reveal the answer.

1 Cards 1–10 are placed in a hat. What is P(card > 7)?

Favourables: {8, 9, 10} = 3. Total = 10.P(card > 7) = 3/10 = 0.3
2 A letter is picked at random from AUSTRALIA. What is P(vowel)?

AUSTRALIA = A-U-S-T-R-A-L-I-A (9 letters). Vowels: A, U, A, I, A = 5 vowels.P(vowel) = 5/9 ≈ 0.556
3 What is P(not drawing a king) from a standard 52-card deck?

P(king) = 4/52 = 1/13. P(not king) = 1 − 1/13 = 12/13.P(not king) = 12/13 ≈ 0.923
4 You roll a fair die 5 times and get a 6 every time. What is P(6) on the next roll?

Still 1/6. Each roll is independent. Past results don't affect future rolls. This common mistake is called the gambler's fallacy.P(6) = 1/6 always — past results don't matter

Complete in your workbook.

Quick Check · 5 questions

A fair die is rolled. What is P(even number)?

+10 XP

P(A) = 3/8. What is P(not A)?

+10 XP

A bag has one of each: red, blue, green, yellow, purple marble. P(yellow) =

+10 XP

What is the probability of rolling a 7 on a standard die?

+10 XP

A bag has 6 red and 9 blue balls. What is P(not blue)?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Medium 3 MARKS

Q6. A bag contains 5 red, 4 blue, 2 yellow and 1 green marble. One is drawn at random. Find: (a) P(red), (b) P(not red), (c) P(yellow or green).

Answer in your workbook.

Understand Easy 2 MARKS

Q7. A letter is chosen at random from PROBABILITY. (a) How many letters in total? (b) What is P(B)?

Answer in your workbook.

Reason Hard 4 MARKS

Q8. The probability that a student passes a test is 0.72. (a) What is P(fail)? (b) In a class of 25, how many would you expect to pass? (c) Explain why your answer to (b) is an expected value, not a guarantee.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — Evens {2,4,6} = 3. P = 3/6 = 1/2.

2. B — P(not A) = 1 − 3/8 = 5/8.

3. A — 1 yellow from 5 = 1/5.

4. D — P = 0; 7 is impossible on a standard die.

5. C — P(not blue) = 6/15 = 2/5.

Show Your Working Model Answers

Q6 (3 marks): Total = 5+4+2+1 = 12. (a) P(red) = 5/12 [1]. (b) P(not red) = 1 − 5/12 = 7/12 [1]. (c) P(yellow or green) = (2+1)/12 = 3/12 = 1/4 [1].

Q7 (2 marks): PROBABILITY has 11 letters: P-R-O-B-A-B-I-L-I-T-Y [1]. B appears twice, so P(B) = 2/11 [1].

Q8 (4 marks): (a) P(fail) = 1 − 0.72 = 0.28 [1]. (b) Expected = 0.72 × 25 = 18 students [1]. (c) Probability is a long-run prediction. In any single class the actual number could be 16, 17, 19 or 20. The expected value tells us what happens on average, not what must happen in any one case [2].

Stretch Challenge · +25 XP, +10 coins

The Unknown Bag

A bag contains only red and blue marbles. P(red) = 3/7. (a) What is P(blue)? (b) If there are 21 marbles total, how many are red? How many are blue? (c) A marble is drawn and NOT put back. Now what is P(red) for the second draw if the first was red?

Reveal solution

(a) P(blue) = 1 − 3/7 = 4/7. (b) Red = 3/7 × 21 = 9; Blue = 4/7 × 21 = 12. (c) After drawing one red (without replacement): 8 red left, 12 blue, 20 total. P(red) = 8/20 = 2/5. This previews Lesson 19 — without replacement changes the probabilities!

Quick Review

Formula

P = favourable ÷ total outcomes

Equally likely

Must verify before using the formula

Complement

P(A') = 1 − P(A) — use for "at least"

Scale 0 to 1

P = 0: impossible. P = 1: certain.

Sum = 1

All probabilities for all outcomes add to 1

Sample space first

List all outcomes before counting

Interactive: Probability Scale Sorter

Place events on the probability scale from 0 (impossible) to 1 (certain). Build your intuition for where everyday events belong.

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