10.4 Theoretical & Experimental Probability

10.4 Theoretical & Experimental Probability

Skills 8 and 9 of Competency 4 pair naturally: skill 8 computes what should happen, skill 9 compares that with what did happen and uses the result to predict.

The two probabilities

Theoretical probability = (number of favorable outcomes) / (number of equally likely outcomes) Experimental probability = (number of successes) / (number of trials)

A standard six-sided die: P(rolling a 4) = 1/6 theoretically. If 300 rolls produce 58 fours, the experimental probability is 58/300 ≈ 0.193, close to 1/6 ≈ 0.167.

Every probability satisfies 0 ≤ P ≤ 1. A probability of 0 means impossible; 1 means certain. Any answer outside that range is wrong by inspection — a useful last check.

The theoretical formula requires equally likely outcomes. For a fair die each face is equally likely, but for the sum of two dice the eleven possible sums are not — there are six ways to roll a 7 and only one to roll a 12, so P(7) = 6/36 and P(12) = 1/36.

The Law of Large Numbers

As the number of trials increases, the experimental probability tends to approach the theoretical probability.

Ten coin flips can easily produce 7 heads (70%); ten thousand flips will land far closer to 50%. This explains why small-sample experimental results deviate from theory without indicating anything is wrong.

[!WARNING] The Law of Large Numbers does not mean outcomes "even out" to compensate for past results. After five heads in a row, the probability of heads on the sixth flip is still 1/2 — the coin has no memory. Believing otherwise is the gambler's fallacy, and diagnosing it in student reasoning is a Competency 5 crossover item.

Predicting counts from probability

Multiply the probability by the number of trials.

A spinner is 40% red. In 250 spins, expect 0.40 × 250 = 100 red outcomes. A bag holds 3 blue and 9 yellow marbles. Drawing with replacement 400 times, expect (3/12)(400) = 100 blue draws.

Items reverse this too: if 45 of 180 trials were successes, the experimental probability is 45/180 = 0.25, and you would predict 0.25 × 1,000 = 250 successes in 1,000 trials.

Compound events

Independent events — the outcome of one does not affect the other.

P(A and B) = P(A) · P(B)

Rolling a 5 and then a 5: (1/6)(1/6) = 1/36. Flipping three heads in a row: (1/2)³ = 1/8. Drawing a red marble twice with replacement from a bag of 4 red and 6 blue: (4/10)(4/10) = 4/25.

Dependent events — the first outcome changes the second probability. This is the without replacement case.

Drawing two red marbles without replacement from 4 red and 6 blue: First draw: 4/10. Second draw: only 3 red remain out of 9 total, so 3/9. P = (4/10)(3/9) = 12/90 = 2/15.

Reducing both the numerator and the denominator is the step candidates miss. After removing a red marble, both the red count and the total drop by one.

Mutually exclusive events cannot occur together.

P(A or B) = P(A) + P(B) when A and B are mutually exclusive.

Rolling a 2 or a 5: 1/6 + 1/6 = 1/3.

When events can overlap, subtract the double-counted intersection:

P(A or B) = P(A) + P(B) − P(A and B)

Drawing a heart or a face card from a standard 52-card deck: 13/52 + 12/52 − 3/52 = 22/52 = 11/26. The three cards subtracted are the jack, queen, and king of hearts, which were counted in both groups.

The complement rule

P(not A) = 1 − P(A)

This is the fastest route for "at least one" questions, because "at least one" is the complement of "none."

A basketball player makes 70% of free throws. What is the probability she makes at least one of her next three? P(misses all three) = (0.30)³ = 0.027. P(at least one) = 1 − 0.027 = 0.973.

Computing "at least one" directly requires adding the cases for exactly one, exactly two, and exactly three — far more work for the same answer.

Simple versus repeated experiments

The blueprint distinguishes simple experiments — one trial, such as a single roll of a die — from repeated experiments, in which the same trial is performed several times and you ask about the combined outcome.

A repeated experiment with independent trials multiplies the per-trial probability once for each repetition:

P(a specific outcome on every one of n independent trials) = P(one trial)ⁿ

Rolling a die four times, P(a 6 every time) = (1/6)⁴ = 1/1,296. Drawing with replacement from a bag of 4 red and 6 blue, P(red on all three draws) = (0.4)³ = 0.064.

Repeated experiments are also where the complement rule earns its keep, because "at least one success in n trials" is always 1 − P(no successes)ⁿ. And note that repetition with replacement keeps trials independent, while repetition without replacement makes them dependent, which is the distinction developed above.

Counting outcomes

The Fundamental Counting Principle: if one choice has m options and another has n, the pair has m × n possible outcomes.

A menu with 4 entrées, 3 sides, and 2 drinks offers 4 × 3 × 2 = 24 meals. Two dice have 6 × 6 = 36 equally likely outcomes.

Organized lists, tables, and tree diagrams all implement this principle and are the standard middle-grades tools for making the sample space visible.

Comparing theoretical and experimental results

Skill 9 asks for the comparison and the conclusion drawn from it.

A die is rolled 600 times and produces 142 sixes. Theoretically we expect 100. Experimental probability = 142/600 ≈ 0.237 versus theoretical 0.167.

With 600 trials, a gap that large is substantial evidence that the die may be unfair or weighted. The reasoning that makes this a conclusion rather than a guess is the Law of Large Numbers: at 600 trials, the experimental value should be close to the theoretical one, so a persistent gap points to a flawed assumption. With only 12 rolls, by contrast, 3 sixes instead of 2 would be entirely unremarkable — sample size determines how much weight the discrepancy carries.