Lesson 17: Probability, Counting, and the Binomial Distribution

Probability begins with counting: most probabilities for equally likely outcomes are a ratio of two counts. This lesson builds the counting toolkit (permutations and combinations), then the binomial distribution — the probability model for repeated independent yes/no trials, and one of the most-used distributions in science and industry.

Counting: permutations and combinations

If order matters, use permutations: P(n, r) = n!/(n − r)!. If order does not matter, use combinations: C(n, r) = n!/(r!(n − r)!). Worked examples. Gold–silver–bronze medals among 8 runners: order matters, P(8, 3) = 8 × 7 × 6 = 336. A committee of 4 from 10 people: order irrelevant, C(10, 4) = 210. With restrictions: choose 5 from 6 seniors and 8 juniors with exactly 3 seniors: C(6, 3)·C(8, 2) = 20 × 28 = 560 — choose each subgroup independently, then multiply (the multiplication principle). "At least one" problems almost always route through the complement: P(at least one) = 1 − P(none). The probability of at least one six in four die rolls is 1 − (5/6)^4 = 1 − 625/1296 = 671/1296 ≈ 0.518 — barely better than a coin flip, which surprises most people and is why the complement method matters.

The binomial setting

A binomial experiment has: a fixed number n of independent trials, two outcomes per trial (success/failure), and a constant success probability p. The random variable X counts successes, and

P(X = k) = C(n, k) · p^k · (1 − p)^(n − k),   mean μ = np,   variance σ^2 = np(1 − p)

Worked example: a 10-question quiz on which a prepared guesser has p = 0.3 on each question. P(exactly 3 correct) = C(10, 3)(0.3)^3(0.7)^7 = 120 × 0.027 × 0.0824 ≈ 0.267. The mean is np = 3 — and note the most likely single value (3) still has probability only about 0.27: in binomial distributions, "expected" does not mean "probable," it means the long-run average. The standard deviation sqrt(2.1) ≈ 1.45 sets the natural spread: scores from 0 to 6 are all unremarkable.

Modelling: quality control and expected value

Quality control. A factory has a 5% defect rate. An inspector samples 20 items: X ~ Binomial(20, 0.05). P(no defects) = (0.95)^20 ≈ 0.358, so P(at least one defect) ≈ 0.642 — finding a defect in the sample is more likely than not, despite the low rate, because 20 trials accumulate risk. The expected number is np = 1 defect per sample, which is how the factory sets its inspection capacity. Expected value generalises the mean: a carnival game costs $2 to roll a die and pays the face value in dollars. The expected payout is (1 + 2 + 3 + 4 + 5 + 6)/6 = $3.50, so the expected net gain is +$1.50 per play — a game no carnival would actually offer, which is the point of computing it.

Common mistakes

• Using permutations for committees. If swapping two selected people produces the same outcome, it is a combination. Medals distinguish positions (permutation); committees do not.
• "At least one" computed head-on. Summing P(1) + P(2) + P(3) + P(4) for the dice problem works but wastes time and invites arithmetic errors; the complement is one term.
• Swapping p^k and (1 − p)^(n − k). The exponent on p is the number of *successes* k; on (1 − p) it is the number of failures n − k. In P(X = 3) above: (0.3)^3(0.7)^7 — exponents 3 and 7 sum to n = 10 ✓; always check they do.
• Assuming independence carelessly. Drawing cards without replacement is not binomial (p changes each draw). Binomial needs independent trials with constant p — sampling from a huge population approximates it, sampling from a small deck does not.
• Reading the mean as a guarantee. μ = np = 3 does not promise 3 successes in 10 trials; P(X = 3) ≈ 0.267 means it fails to happen about 73% of the time. Means describe averages over many repetitions.

Exam Experience

1. The probability of exactly 3 heads in 5 fair coin tosses is

• (A) 3/32
• (B) 5/16
• (C) 1/2
• (D) 3/5

Answer: (B). C(5, 3)(1/2)^3(1/2)^2 = 10/32 = 5/16. (A) uses 3 favourable outcomes instead of C(5, 3) = 10 — the 10 distinct arrangements of 3 heads and 2 tails count, not just the tally.

2. The number of different 4-person committees that can be formed from 12 students is

• (A) 11,880
• (B) 495
• (C) 48
• (D) 20,736

Answer: (B). C(12, 4) = 12!/(4!·8!) = 495. (A) is P(12, 4) — order counted when it should not be; (D) is 12^4, counting with repetition and order.

3. For X ~ Binomial(50, 0.2), the mean and standard deviation are

• (A) μ = 10, σ = 8
• (B) μ = 10, σ = 2.83
• (C) μ = 40, σ = 2.83
• (D) μ = 2, σ = 10

Answer: (B). μ = np = 10; σ = sqrt(np(1 − p)) = sqrt(50 × 0.2 × 0.8) = sqrt(8) ≈ 2.83. (A) reports the variance 8 as the standard deviation — the missing square root is the trap.

Constructed response. A production line has a defect rate of 5%. An inspector samples 20 items independently, and X counts the defects. (a) State why X can be modelled as binomial. (b) Find P(X = 0) and P(X ≥ 1). (c) Find P(X = 1) and hence P(X ≤ 1). (d) Compute the expected number of defects and interpret it for the inspector.

Scoring guide. (a) 1 point: fixed n = 20 trials, each item defective or not, constant p = 0.05, independent sampling. (b) 2 points: P(X = 0) = (0.95)^20 ≈ 0.358 — 1 point; P(X ≥ 1) = 1 − 0.358 = 0.642 — 1 point. (c) 2 points: P(X = 1) = C(20, 1)(0.05)(0.95)^19 = 20 × 0.05 × 0.3774 ≈ 0.377 — 1 point; P(X ≤ 1) ≈ 0.358 + 0.377 = 0.736 — 1 point. (d) 1 point: μ = np = 20 × 0.05 = 1 defect per sample on average over many samples — not a promise of exactly one in any given sample.

Going deeper: conditional probability and the base-rate trap

Conditional probability P(A|B) = P(A and B)/P(B) answers "given that B happened, how likely is A now?" Its most consequential application is medical testing, where the base rate dominates intuition. Worked example: a disease affects 1% of a population. A test detects it 99% of the time when present (sensitivity) and falsely flags 5% of healthy people. You test positive — what is P(disease | positive)? Out of 10,000 people: 100 have the disease and 99 test positive; 9,900 are healthy and 5% × 9,900 = 495 falsely test positive. So P = 99/(99 + 495) = 99/594 = 1/6 ≈ 16.7%. Despite a "99% accurate" test, five of six positives are false — because the disease is rare, the small false-positive rate applied to the huge healthy group swamps the true positives. This base-rate fallacy misleads doctors, jurors and headline-writers; the frequency table (out of 10,000) is the antidote, and it is pure counting — Lesson 17's tools in a white coat.

Two vocabulary guards complete the topic. Independent events (one does not change the other's probability: separate coin tosses) are routinely confused with mutually exclusive events (cannot co-occur: one toss landing heads vs tails) — in fact mutually exclusive events with positive probabilities are maximally *dependent*, since one occurring forces the other to probability zero. And expected value extends to the binomial: E[X] = np is the same averaging idea as the carnival game, now distributed over counts — one concept, many costumes.

Workshop: two dice, the birthday problem and waiting times

Two dice. The 36 equally likely outcomes give the sum distribution: counts 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for sums 2 through 12. So P(sum = 7) = 6/36 = 1/6 — the most likely sum — and the expected sum is 7 by symmetry (the distribution is symmetric about 7), a result casinos price their games against. The table *is* the counting: each count is a small combination problem (sum 7 from (1,6), (2,5), (3,4) and their swaps).

The birthday problem. In a class of 23, the probability that at least two students share a birthday is about 0.507 — better than even, despite 365 possible birthdays. The complement does the work: P(no match) = (365/365)(364/365)(363/365)...(343/365) ≈ 0.493, since each new student must dodge all birthdays already taken. The intuition repair: you are not asking whether someone matches *you* (22 chances at 1/365) but whether *any pair* matches — and there are C(23, 2) = 253 pairs. Waiting times: the expected number of rolls to see a first six is 1/p = 6 — the geometric distribution's mean, and the reason "due for a six" thinking fails: the die has no memory, whatever the wait so far.

Final drills: complements and a small binomial

Complement fluency. P(at least one head in 3 fair tosses) = 1 − P(no heads) = 1 − (1/2)^3 = 7/8. The complement converts a 7-case enumeration into one subtraction — the same move as the dice and defect problems, and the single most time-saving reflex in elementary probability. A small binomial, complete: for X ~ Binomial(4, 1/2) (four coin tosses, counting heads), P(X = 2) = C(4, 2)(1/2)^2(1/2)^2 = 6/16 = 3/8. The full distribution 1/16, 4/16, 6/16, 4/16, 1/16 is row 4 of Pascal's triangle scaled by 1/16 — the binomial theorem of Lesson 10 *is* the binomial distribution when p = 1/2, and the general case is the same expansion with weighted terms. Odds, translated: bookmakers quote odds a-to-b in favour, meaning probability a/(a + b); 3-to-1 in favour is probability 3/4, and a fair die's prime faces (2, 3, 5) give probability 3/6 = 1/2, i.e. odds of 1-to-1 — "evens." Convert on sight; the two dialects describe one number.