Mathematics • Year 9 • Unit 4 • Lesson 13
Introduction to Probability
Build fluency with the basic probability formula P(E) = favourable ÷ total, sample spaces and the complement rule P(E') = 1 − P(E), from a worked example through guided practice to eight independent problems.
1. I do, fully worked example
Read every step. The reason on the right tells you why, not just what.
Problem. A bag holds 3 red, 5 blue and 2 green marbles. A marble is drawn at random. Find P(blue) and P(not green) and check the complement rule.
Step 1, Count the total outcomes.
Total = 3 + 5 + 2 = 10 marbles.
Reason: the denominator in P(E) is always the total number of equally likely outcomes.
Step 2, Count the favourable outcomes for "blue".
Favourable = 5 blue marbles.
Step 3, Apply the probability formula.
P(blue) = favourable ÷ total = 5 ÷ 10 = 1/2.
Reason: simplify the fraction when you can.
Step 4, Find P(green) directly.
P(green) = 2 ÷ 10 = 1/5.
Step 5, Use the complement rule for P(not green).
P(not green) = 1 − P(green) = 1 − 1/5 = 4/5.
Reason: P(E) + P(E') = 1 always, so P(E') = 1 − P(E).
Step 6, Check directly.
"Not green" means red or blue, so favourable = 3 + 5 = 8. P(not green) = 8 ÷ 10 = 4/5 ✓.
Answer: P(blue) = 1/2, P(not green) = 4/5.
2. We do, fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 5 marks
Problem. A bag holds 4 red, 6 blue and 10 green marbles. Find P(red), P(blue), P(green) and P(not red).
Step 1, Total outcomes: total = 4 + 6 + 10 = ______ .
Step 2, P(red):
Favourable = ______ . P(red) = ______ ÷ ______ = ______ (simplified).
Step 3, P(blue):
Favourable = ______ . P(blue) = ______ ÷ ______ = ______ (simplified).
Step 4, P(green):
Favourable = ______ . P(green) = ______ ÷ ______ = ______ (simplified).
Step 5, P(not red) by complement:
P(not red) = 1 − P(red) = 1 − ______ = ______ .
Step 6, Check sum:
P(red) + P(blue) + P(green) = ______ + ______ + ______ = ______ . Should be 1.
3. You do, independent practice
Show working under each problem. Foundation = single P(E); Standard = combine ideas; Extension = sample-space or complement reasoning.
Foundation, single probability
3.1 A fair die is rolled. Find P(rolling a 4). 1 mark
3.2 A bag has 3 red and 7 blue marbles. Find P(red). 1 mark
3.3 A standard 52-card deck. Find P(heart). 1 mark
3.4 If P(rain tomorrow) = 0.3, find P(no rain tomorrow). 1 mark
Standard, combine ideas
3.5 A spinner has 8 equal sectors: 3 red, 3 blue, 2 green. Find P(red), P(green) and P(not green). 3 marks
3.6 A fair die is rolled. Find P(even number) and P(number greater than 4). 3 marks
Extension, sample space / complement
3.7 Two fair coins are tossed. List the sample space and find P(at least one head). Use the complement to double-check your answer. 3 marks
3.8 A letter is chosen at random from the word PROBABILITY. Find P(vowel) and P(consonant), and verify P(vowel) + P(consonant) = 1. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (4R / 6B / 10G)
Total = 20.
P(red) = 4 ÷ 20 = 1/5. P(blue) = 6 ÷ 20 = 3/10. P(green) = 10 ÷ 20 = 1/2.
P(not red) = 1 − 1/5 = 4/5.
Check: 1/5 + 3/10 + 1/2 = 2/10 + 3/10 + 5/10 = 10/10 = 1 ✓.
3.1, P(rolling a 4)
Favourable = 1, total = 6. P(4) = 1/6.
3.2, P(red) from 3R / 7B
Total = 10. P(red) = 3 ÷ 10 = 3/10.
3.3, P(heart)
A deck has 4 suits of 13. P(heart) = 13 ÷ 52 = 1/4.
3.4, P(no rain)
P(no rain) = 1 − 0.3 = 0.7.
3.5, Spinner with 3R / 3B / 2G out of 8
P(red) = 3/8. P(green) = 2/8 = 1/4. P(not green) = 1 − 1/4 = 3/4.
3.6, Fair die
Even = {2, 4, 6}: P(even) = 3/6 = 1/2. Greater than 4 = {5, 6}: P(>4) = 2/6 = 1/3.
3.7, Two fair coins
Sample space = {HH, HT, TH, TT}, 4 equally likely outcomes.
"At least one head" = {HH, HT, TH}: P = 3/4.
Complement check: P(no heads) = P(TT) = 1/4, so P(at least one H) = 1 − 1/4 = 3/4 ✓.
3.8, PROBABILITY
Total letters = 11. Vowels: O, A, I, I = 4. Consonants: P, R, B, B, L, T, Y = 7.
P(vowel) = 4/11. P(consonant) = 7/11.
Check: 4/11 + 7/11 = 11/11 = 1 ✓.