Mathematics • Year 8 • Unit 4 • Lesson 14
Introduction to Probability
Build fluency with the probability formula, sample spaces, and complementary events. One worked example, one guided fill-in with blanks, then eight independent problems from quick recall to combined experiments.
1. I do — fully worked example
Read every line. Each step has a short reason.
Problem. A bag contains 3 red, 4 blue, and 5 green marbles. A marble is drawn at random. Find P(blue) and P(not blue).
Step 1 — Write the sample space size (total outcomes).
Total marbles = 3 + 4 + 5 = 12
Reason: every marble is equally likely to be drawn, so the total goes in the denominator.
Step 2 — Count favourable outcomes for the event.
Favourable for "blue" = 4 (there are 4 blue marbles)
Reason: favourable outcomes are the ones that satisfy the event we're interested in.
Step 3 — Apply the probability formula.
P(blue) = favourable ÷ total = 4 ÷ 12 = 1/3
Reason: simplify the fraction when possible. 4/12 = 1/3.
Step 4 — Use the complement rule for "not blue".
P(not blue) = 1 − P(blue) = 1 − 1/3 = 2/3
Reason: an event and its complement together cover everything, so they sum to 1.
Answer: P(blue) = 1/3 ≈ 0.33 = 33.3%. P(not blue) = 2/3 ≈ 0.67 = 66.7%.
2. We do — fill in the missing steps
Same shape as Section 1, but the working is faded. Fill in each blank. 4 marks
Problem. A standard 6-sided die is rolled. Find P(rolling a number greater than 4) and P(not rolling a number greater than 4).
Step 1 — Total outcomes:
Sample space = {1, 2, 3, 4, 5, 6}, total = ______
Step 2 — Favourable outcomes (greater than 4):
Favourable = {______, ______}, count = ______
Step 3 — Apply the probability formula:
P(greater than 4) = ______ ÷ ______ = ______ (simplify)
Step 4 — Use the complement rule:
P(not greater than 4) = 1 − ______ = ______
3. You do — independent practice
Show your working in the space under each problem. Foundation → Standard → Extension.
Foundation — quick probability calculations
3.1 Find P(heads) when flipping a fair coin. 1 mark
3.2 A bag has 5 red and 3 yellow balls. Find P(red) as a fraction. 1 mark
3.3 Find P(rolling a 3) on a standard fair die. 1 mark
3.4 Define sample space in one short sentence and give one example. 1 mark
Standard — formula with bigger numbers and complements
3.5 A standard deck of 52 cards. Find (a) P(King) (as a fraction), (b) P(red card) (as a fraction). 2 marks
3.6 The probability of rain tomorrow is 0.4. Use the complement rule to find the probability of NO rain. 2 marks
Extension — combined experiments and reasoning
3.7 Two standard dice are rolled together. (a) How many outcomes are in the sample space? (b) Find P(rolling double 6). 2 marks
3.8 A bag has 4 red, 6 blue, and 2 green marbles. A marble is drawn at random and replaced. If this is repeated 60 times, how many BLUE marbles should you expect? Show the formula: expected = trials × P(event). 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (greater than 4 on a die)
Step 1: Total = 6.
Step 2: Favourable = {5, 6}, count = 2.
Step 3: P(greater than 4) = 2 ÷ 6 = 1/3.
Step 4: P(not greater than 4) = 1 − 1/3 = 2/3.
3.1 — P(heads)
Total = 2, favourable = 1. P(heads) = 1 ÷ 2 = 1/2.
3.2 — P(red) from bag
Total = 5 + 3 = 8. P(red) = 5 ÷ 8 = 5/8.
3.3 — P(rolling a 3)
Total = 6, favourable = 1 (just one "3"). P(3) = 1/6.
3.4 — Define sample space
The sample space is the complete set of all possible outcomes of an experiment. Example: for a coin flip, sample space = {Heads, Tails}.
3.5 — Standard deck
(a) P(King) = 4 ÷ 52 = 1/13. (b) P(red card) = 26 ÷ 52 = 1/2 (red = hearts + diamonds = 13 + 13 = 26).
3.6 — Complement: no rain
P(no rain) = 1 − P(rain) = 1 − 0.4 = 0.6 (or 60%).
3.7 — Two dice
(a) Each die has 6 outcomes, so total = 6 × 6 = 36 outcomes.
(b) Only one outcome is (6, 6). P(double 6) = 1/36.
3.8 — Expected blue marbles
Total marbles = 4 + 6 + 2 = 12. P(blue) = 6 ÷ 12 = 1/2. Expected blue in 60 draws = 60 × 1/2 = 30 blue marbles.