Mathematics • Year 7 • Unit 4 • Lesson 17
Theoretical Probability
Build fluency with the four-step routine: list the sample space → check equally likely → count favourables → apply P = f/n. Then use the complement rule P(A') = 1 − P(A) as a shortcut.
1. I do — fully worked example
Read every line — every step is named.
Problem. A standard die is rolled. Find P(rolling a prime number).
Step 1 — List the sample space.
{1, 2, 3, 4, 5, 6} → n = 6 total outcomes
Reason: writing out all outcomes is what stops you from forgetting one.
Step 2 — Check equally likely.
A fair die → every face has the same chance → P = f/n applies. ✓
Step 3 — Count favourable outcomes.
Primes on a die: {2, 3, 5} → f = 3
Reason: 1 is NOT prime; 4 and 6 are composite.
Step 4 — Apply the formula and simplify.
P(prime) = 3 ÷ 6 = 1/2 = 0.5 = 50%
Answer: P(prime) = 1/2.
2. We do — fill in the missing steps
A bag has 4 red, 3 blue and 5 green marbles. One marble is drawn at random. Find P(not green). Fill in each blank. 4 marks
Step 1 — Find the total.
Total marbles = 4 + 3 + 5 = _______
Step 2 — Check equally likely.
Well-mixed bag → all marbles are _______________ likely.
Step 3 — Count favourables (not green).
Not green = red + blue = ___ + ___ = _______
Step 4 — Apply the formula AND check with the complement.
Direct: P(not green) = ___ ÷ ___ = _______
Complement: 1 − P(green) = 1 − 5/12 = _______
3. You do — independent practice
Show sample space → count favourables → answer in lowest terms (and as a decimal).
Foundation — apply P = f/n
3.1 A die is rolled. Find P(rolling a 4). 1 mark
3.2 Cards numbered 1–10 are placed in a hat. One card is drawn. Find P(card > 7). 1 mark
3.3 A bag has 5 red and 7 blue marbles. Find P(red). 2 marks
3.4 A standard 52-card deck. Find P(drawing a heart). 2 marks
Standard — use the complement rule
3.5 A die is rolled. Find P(not 6) using the complement rule. 2 marks
3.6 Standard 52-card deck. Use the complement rule to find P(not a king). 2 marks
Extension — push your thinking
3.7 A letter is picked at random from the word AUSTRALIA. (i) Find P(vowel). (ii) Find P(letter A). (iii) Find P(consonant) using the complement rule. 3 marks
3.8 Two fair coins are flipped. (i) List the sample space. (ii) Find P(at least one head) using the complement rule. (iii) Verify by listing favourables directly. 3 marks
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What I'll revisit before next class:
Section 2 — Marble bag (We do)
Step 1: Total = 4 + 3 + 5 = 12.
Step 2: All marbles are equally likely.
Step 3: Not green = 4 + 3 = 7.
Step 4: Direct: 7 ÷ 12 = 7/12. Complement: 1 − 5/12 = 7/12. ✓
3.1 — P(rolling a 4)
Sample space {1, 2, 3, 4, 5, 6}, n = 6, favourable = 1. P(4) = 1/6 ≈ 0.167.
3.2 — P(card > 7) from 1–10
Sample space {1, …, 10}, n = 10. Favourables {8, 9, 10}, f = 3. P = 3/10 = 0.3.
3.3 — P(red), 5 red and 7 blue
Total = 12. P(red) = 5/12 ≈ 0.417.
3.4 — P(heart) from 52-card deck
52 cards, 13 hearts. P(heart) = 13/52 = 1/4 = 0.25.
3.5 — P(not 6) on a die
P(6) = 1/6, so P(not 6) = 1 − 1/6 = 5/6 ≈ 0.833.
3.6 — P(not king), 52-card deck
P(king) = 4/52 = 1/13. P(not king) = 1 − 1/13 = 12/13 ≈ 0.923.
3.7 — AUSTRALIA (A-U-S-T-R-A-L-I-A)
9 letters total. (i) Vowels A, U, A, I, A = 5. P(vowel) = 5/9 ≈ 0.556. (ii) Letter A appears 3 times. P(A) = 3/9 = 1/3. (iii) Consonants = 9 − 5 = 4. P(consonant) = 1 − 5/9 = 4/9 ≈ 0.444.
3.8 — Two coins, P(at least one head)
(i) Sample space {HH, HT, TH, TT}, n = 4. (ii) Complement = "no heads" = TT only; P(TT) = 1/4. So P(at least one head) = 1 − 1/4 = 3/4 = 0.75. (iii) Direct check: favourables {HH, HT, TH} = 3 out of 4 → 3/4. ✓