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.