Mathematics • Year 9 • Unit 4 • Lesson 16
Experimental and Theoretical Probability
Build fluency with the two probability formulas: theoretical P(E) = favourable ÷ total, and experimental P(E) ≈ occurrences ÷ trials. One worked example, one guided example with blanks, then eight graduated practice problems.
1. I do — fully worked example
Read every line. Each step shows what is being calculated and why.
Problem. A coin is tossed 200 times. (a) How many heads would you expect? (b) If 95 heads actually occur, find the experimental P(head). (c) Comment on whether this suggests bias.
Step 1 — Theoretical first: what does fair predict?
Theoretical P(head) = 1 ÷ 2 = 0.5 (two equally likely outcomes, one is "head").
Reason: theoretical probability = favourable ÷ total possible outcomes, assuming fairness.
Step 2 — Expected number of heads in 200 tosses.
Expected = 200 × 0.5 = 100 heads.
Reason: expected count = (number of trials) × (theoretical probability).
Step 3 — Experimental P(head) from the actual 95 heads.
Experimental P(head) = 95 ÷ 200 = 0.475.
Reason: experimental P = (occurrences of event) ÷ (total trials). This is also called relative frequency.
Step 4 — Comment on bias.
0.475 is close to 0.5. Difference of 5 heads out of 200 is small and easily caused by random variation. No clear evidence of bias.
Answer: (a) 100 heads expected; (b) experimental P(head) = 0.475; (c) likely just random variation — not biased.
2. We do — fill in the missing steps
Same structure as Section 1. Fill the blanks as you go. 4 marks
Problem. A spinner has 4 equal sections (red, blue, green, yellow). In 80 spins, red appears 25 times. (a) Find theoretical P(red). (b) Find experimental P(red). (c) Compare.
Step 1 — Theoretical P(red). Four equally likely sections, one is red.
Theoretical P(red) = ____ ÷ ____ = ______
Step 2 — Expected reds in 80 spins.
Expected = 80 × ______ = ______ reds.
Step 3 — Experimental P(red).
Experimental P(red) = ____ ÷ ____ = ______
Step 4 — Compare and explain.
Experimental is ______ than theoretical. With only ____ spins, this difference is most likely due to ____________________.
3. You do — independent practice
Show working under each problem. Foundation = single rule; standard = combine two ideas; extension = compare or evaluate.
Foundation — single calculation
3.1 A fair die is rolled. Find the theoretical probability of rolling a 5. 1 mark
3.2 A coin is tossed 50 times. Heads come up 22 times. Find the experimental P(head) as a decimal. 1 mark
3.3 A die is rolled 60 times. How many 6s do you expect? 1 mark
3.4 A spinner has 5 equal sections. What is the theoretical probability of landing on any one section? Express as a fraction and a decimal. 1 mark
Standard — combine two ideas
3.5 A die is rolled 90 times. (a) How many of each number are expected? (b) If the number 4 actually appears 20 times, find the experimental P(4). 2 marks
3.6 A bag has unknown marbles. After 100 draws (with replacement), 40 are red. (a) Estimate P(red). (b) If the bag holds 250 marbles, roughly how many are red? 2 marks
Extension — compare and reason
3.7 A coin is tossed three times: 10, 100, 1000 tosses. The heads counts are 7, 54 and 503 respectively. Calculate experimental P(head) for each, then explain in one sentence which result is most trustworthy and why. 3 marks
3.8 A die is rolled 600 times. Theoretical P(6) = 1/6. The number 6 actually appears 145 times. (a) What was expected? (b) By how much does experimental P differ from theoretical? (c) Is this difference small enough to ignore? Explain. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded spinner)
Step 1: theoretical P(red) = 1 ÷ 4 = 0.25.
Step 2: expected = 80 × 0.25 = 20 reds.
Step 3: experimental P(red) = 25 ÷ 80 = 0.3125.
Step 4: experimental is higher than theoretical. With only 80 spins, this difference is most likely due to random variation (not enough trials).
3.1 — Theoretical P(5) on a die
One favourable outcome (the 5), six possible outcomes. P(5) = 1/6 ≈ 0.167.
3.2 — Experimental P(head)
P(head) = 22 ÷ 50 = 0.44.
3.3 — Expected 6s in 60 rolls
Expected = 60 × (1/6) = 10 sixes.
3.4 — 5-section spinner
P(any section) = 1/5 = 0.2.
3.5 — Die rolled 90 times
(a) Expected of each number = 90 ÷ 6 = 15.
(b) Experimental P(4) = 20 ÷ 90 = 2/9 ≈ 0.222.
3.6 — Marble bag
(a) Experimental P(red) ≈ 40 ÷ 100 = 0.4.
(b) About 0.4 × 250 = 100 red marbles.
3.7 — Coin tosses 10 / 100 / 1000
P(head) = 7/10 = 0.7; 54/100 = 0.54; 503/1000 = 0.503.
The 1000-toss result is most trustworthy because more trials reduce random variation — by the law of large numbers, experimental probability approaches the theoretical 0.5 as trials grow.
3.8 — Die rolled 600 times
(a) Expected = 600 ÷ 6 = 100 sixes.
(b) Experimental P(6) = 145 ÷ 600 ≈ 0.2417. Theoretical = 1/6 ≈ 0.1667. Difference ≈ 0.075.
(c) Probably not negligible — getting 145 sixes when 100 were expected is 45 more than expected. With this many trials, random variation alone is unlikely to cause that big a gap, so the die may be biased and should be tested further.