Probability Experiments — Single Events
Warm-up
Roll a die: P(even) = 3/6 = 1/2. P(greater than 4) = 2/6 = 1/3. P(7) = 0/6 = 0. P(1-6) = 6/6 = 1. Probabilities range from 0 (impossible) to 1 (certain). As a percent: 50%, 33.3%, 0%, 100%. Connect fraction to decimal to percent for each.
Explore
Probability experiment with fractions: spinner with 5 equal sections (2 red, 2 blue, 1 green). Theoretical: P(red)=2/5=0.4=40%, P(blue)=2/5=40%, P(green)=1/5=0.2=20%. Run 30 spins. Calculate experimental probabilities as fractions. Compare to theoretical. Pool class results: 30x number of groups = much larger sample.
Formalize
Independence demonstration: flip a coin 3 times, get 3 heads. What is P(heads) on the 4th flip? Still 1/2. The coin is not due for tails. Show this with data: track what happens after a run of 3 heads in many trial sequences. About half the time, the next flip is heads. The coin truly has no memory.
Probability Experiments — Single Events
Probability scale: place events on a 0-to-1 scale as fractions and decimals. Roll a 7: 0/6=0. Roll a 1 or 2: 2/6=1/3=0.33. Roll any number: 6/6=1. Draw a heart from a deck: 13/52=1/4=0.25. Draw a card over 10: kings, queens, jacks, aces = 16/52=4/13=0.31.
Practice
Students calculate theoretical probabilities for 6 different experiments (as fractions, decimals, and percents), run one experiment with 30 trials, and compare to theoretical. Exit ticket: P(rolling an even number) as a fraction, decimal, and percent.
Exit ticket
Students calculate theoretical probabilities for 6 different experiments (as fractions, decimals, and percents), run one experiment with 30 trials, and compare to theoretical. Exit ticket: P(rolling an even number) as a fraction, decimal, and percent.
Step 1: The upgrade this year: likelihood words (likely, unlikely) become NUMBERS — a probability is a fraction between 0 (impossible) and 1 (certain).
Step 2: Compute from equally-likely outcomes: a standard die → P(roll a 5) = 1/6 (one face among six equal faces). P(even) = 3/6 = 1/2 (faces 2, 4, 6). P(number below 7) = 6/6 = 1 (certain). P(roll a 9) = 0/6 = 0.
Step 3: Place each on the 0-to-1 probability line: 0 … 1/6 … 1/2 … 1. The line unifies last year's word-ranks with this year's fractions — "unlikely" is the neighbourhood below 1/2, now with street addresses.
Step 4: The requirement under the formula, stressed hard: count outcomes ONLY when they're equally likely. "I either win the lottery or I don't — two outcomes, so 1/2" fails the requirement spectacularly. Equal-likelihood is a property of fair dice, drawn slips, balanced spinners — it must be argued, not assumed.
Exit: a bag holds 3 red, 5 blue, 2 green. P(blue)? (5/10 = 1/2.) P(not green)? (8/10 = 4/5.)
Step 1: Before rolling: P(six) = 1/6, so in 60 rolls expect ABOUT 1/6 of 60 = 10 sixes. Write the expected count for every face: 10 each.
Step 2: Roll 60 times (pairs, tally sheet). Sample result: face counts 8, 12, 9, 11, 13, 7.
Step 3: Compare observed vs expected: nobody's face hit exactly 10; everything hovers within a few. Compute each face's experimental probability: 8/60, 12/60 … ≈ 0.13 to 0.22, straddling the theoretical 1/6 ≈ 0.17.
Step 4: Pool the class (600+ rolls): the pooled fractions tighten around 1/6. State the law informally and honestly: experimental probability DRIFTS TOWARD theoretical probability as trials grow — it never owes you exactness on any given day.
Step 5: The two probabilities, named and befriended: THEORETICAL (from counting the symmetric possibilities) and EXPERIMENTAL (from doing the thing). When the two disagree badly and persistently — suspect the die, not the mathematics. That's literally how casinos catch crooked dice.
The game: Spinner 1 shows halves RED/BLUE; Spinner 2 shows thirds RED/BLUE/BLUE. Spin both; Player A scores if the colours MATCH, Player B if they differ. Fair?
Step 1: Gut votes first (usually split), then build the outcome grid: rows = spinner 1 (R, B), columns = spinner 2 (R, B, B — list the two blues separately; they're distinct equally-likely thirds!). Six equally likely cells: RR, RB, RB, BR, BB, BB.
Step 2: Score the cells: matches — RR, BB, BB → 3 cells. Differs — RB, RB, BR → 3 cells. P(match) = 3/6 = 1/2. FAIR — surprisingly.
Step 3: The trap this problem defuses: treating spinner 2 as "red or blue, 50-50" (merging the two blue thirds) — that grid gives 2×2 = 4 cells and the WRONG answer. Outcomes must be split until equally likely — the blue-blue distinction is invisible in the result but essential in the counting.
Step 4: Verify by experiment: 40 plays, tally match/differ — hovers near even ✓.
Exit: change spinner 2 to R/R/B and re-referee the game. (Matches: RR, RR, BB → 3 of 6 — still fair! Then find a spinner pair that ISN'T.)