Binomial & Geometric Distributions
1) Two fair dice rolled 378 times. How often is sum > 8?
2) (¼ + ¾)³ = (¼)³ + 3(¼)²(¾) + 3(¼)(¾)² + (¾)³
Two special discrete distributions
Boil any experiment down to success or failure (profit/loss, guilty/innocent, out/not out). Real life has shades in between, but this yes/no view gives us a mathematical model.
Both distributions come from repeated independent trials where P(success) is the same every trial:
- Binomial → number of successes in a fixed number of independent trials.
- Geometric → number of trials up to and including the first success (could go on forever).
1. The Binomial Distribution
① Building it: roll four fair dice
Let R = number of 6s. So R ∈ {0, 1, 2, 3, 4}. Success = 6 → P(6) = 16, failure X → P(X) = 56.
| r | Ways to get r successes | No. of ways | P(R = r) |
|---|---|---|---|
| 0 | (XXXX) | ⁴C0 = 1 | ⁴C0(16)0(56)4 |
| 1 | (6XXX),(X6XX),(XX6X),(XXX6) | ⁴C1 = 4 | ⁴C1(16)1(56)3 |
| 2 | (66XX),(6X6X),(6XX6),(X66X),(X6X6),(XX66) | ⁴C2 = 6 | ⁴C2(16)2(56)2 |
| 3 | (666X),(66X6),(6X66),(X666) | ⁴C3 = 4 | ⁴C3(16)3(56)1 |
| 4 | (6666) | ⁴C4 = 1 | ⁴C4(16)4(56)0 |
So P(R = r) = 4Cr (16)r (56)4−r — these are exactly the terms of the binomial expansion of (16 + 56)4.
Written with the (nr) notation: P(R = r) = (4r)(16)r(56)4−r.
② Conditions for a binomial distribution (defined by parameters n and p)
- There are n repeated independent trials
- n is finite (fixed in advance)
- Just two outcomes each trial (success / failure)
- P(success) = p is constant every trial
The random variable = number of trials that result in a success. Write X ~ B(n, p).
Example X ~ B(3, p): X ∈ {0,1,2,3}, with q = 1 − p:
P(X=2) = 3p²q P(X=3) = 1p³
③ Pascal's triangle — the coefficients
Coefficients in all binomial expansions are symmetric strings of integers. Arranged in rows they form Pascal's triangle (Blaise Pascal). Each number = sum of the two above it. Row n gives coefficients of (a+b)n.
Regular pentagonal spinner (sectors A, B, C, A, D). Find P(exactly three As in 10 spins).
X = number of As. 10 independent trials, P(A) = 0.4 constant ⇒ X ~ B(10, 0.4). Need 3 successes + 7 failures.
X ~ B(8, 0.7). Find P(X > 6) to 3 s.f.
n = 8, p = 0.7, q = 0.3, X ∈ {0,…,8}. "More than 6" = 7 or 8.
85% of a population has rhesus-positive blood. Find P(fewer than 39 of a random sample of 40 are R+).
X = number R+ in the sample ⇒ X ~ B(40, 0.85). "Fewer than 39" is 0…38 — too long! Use P(A) = 1 − P(A′) and subtract 39 and 40.
X ~ B(n, 0.4) and P(X = 0) < 0.1. Find the least possible n.
First write P(X=0) in terms of n: (n0) × 0.4⁰ × 0.6ⁿ = 0.6ⁿ. Then solve the inequality with logs.
Alternative (trial & improvement): n is an integer, so evaluate 0.6¹, 0.6², … until first < 0.1: 0.6³ = 0.216, 0.6⁴ = 0.1296, 0.6⁵ = 0.07776 ✓ ⇒ n = 5.
- Define X in words, state X ~ B(n, p)
- "exactly r" → one term | "at most / fewer than" → add up | "at least / more than" → often use 1 − …
- Strict (<, >) vs inclusive (≤, ≥): X is an integer, so "> 6" means 7, 8, … and "< 39" means ≤ 38
- Unknown n with an inequality → write in terms of n, use logs (flip sign when dividing by log of number < 1) or trial & improvement
- Don't round partway; give final answers to 3 s.f.
2. Expectation & Variance of the Binomial
Expectation = long-term average value (Ch 6.3) — it is a measure of central tendency; variance / standard deviation measure spread. For a binomial we can get them straight from the parameters n and p.
Check with X ~ B(2, 0.6) (probabilities from the formula):
| x | 0 | 1 | 2 |
|---|---|---|---|
| P(X = x) | 0.16 | 0.48 | 0.36 |
E(X) = np = 2 × 0.6 = 1.2 ✓ (no surprise — Ch 4.1: event A expected n × P(A) times). Surprise: Var(X) = np(1−p) = 2 × 0.6 × 0.4 = 0.48 ✓ too!
X ~ B(12, 0.3). Find the mean, variance and standard deviation.
X ~ B(n, p), E(X) = 12, Var(X) = 7.5. Find a) n and p b) P(X = 11).
3. The Geometric Distribution
① Building it: roll until you get a 6
Let p = P(success), 1 − p = P(failure) (constant). How likely is the first 6 on roll 1? roll 2? roll 3?
- first 6 on 1st roll = p → a success
- first 6 on 2nd roll = (1−p)p → failure then success
- first 6 on 3rd roll = (1−p)²p → two failures then success
X = number of trials up to and including the first success → geometric distribution.
| r | 1 | 2 | 3 | 4 | … | n | … |
|---|---|---|---|---|---|---|---|
| P(X = r) | p | p(1−p) | p(1−p)² | p(1−p)³ | … | p(1−p)ⁿ⁻¹ | … |
These are the terms of a geometric progression (GP): first term p, common ratio (1−p). Sum of all probabilities = sum to infinity:
② Conditions for a geometric distribution (parameter p)
- trials are independent
- trials can be infinite in number
- just two outcomes each trial
- P(success) = p constant
| Binomial B(n, p) | Geometric Geo(p) | |
|---|---|---|
| Number of trials | fixed from the start (n) | repeated until 1st success (could be ∞) |
| X counts | number of successes | trial number of the 1st success |
| Ways to get result | (nr) ways to get r successes | only one way: r−1 failures then a success |
| Values of X | 0, 1, …, n | 1, 2, 3, … |
P(success) = 0.66 each trial. Find P(first success is a) on the 3rd trial b) on or before the 2nd c) after the 3rd).
X = trials up to & including first success ⇒ X ~ Geo(0.66), p = 0.66, 1−p = 0.34.
P(X ≤ r) = 1 − qr P(X > r) = qr
18% of adults wear contact lenses; adults interviewed one at a time. Find P(first lens-wearer is a) one of the first 15 b) not one of the first nine).
X = adults interviewed up to & including first lens-wearer ⇒ X ~ Geo(0.18), q = 0.82.
Biased coin, P(head) = 511. Toss until first head. Find P(tossed a) at least six times b) fewer than eight times).
X ~ Geo(511), q = 611.
4. Mode & Expectation of the Geometric
① Mode
All geometric distributions look alike: P(X = 1) is the biggest, and P(X = r) decreases as r
increases — because the common ratio q = 1 − p < 1:
p > p(1−p) > p(1−p)² > p(1−p)³ > …
| P(X=1) | P(X=2) | P(X=3) | P(X=4) | P(X=5) | |
|---|---|---|---|---|---|
| Geo(0.2) | 0.2 | 0.16 | 0.128 | 0.1024 | 0.08192 |
| Geo(0.7) | 0.7 | 0.21 | 0.063 | 0.0189 | 0.00567 |
② Expectation
Mean = long-term average = Σxp. For Geo(p) it turns out to be the reciprocal of p.
X ∈ {1,2,3,…}, probabilities p, pq, pq², pq³, …
One in four boxes of Zingo cereal contains a free toy. X = number of boxes a child opens, up to and including the one with their first toy. a) Find the mode and E(X). b) Interpret both in context.
b) A child is most likely to find their first toy in the first box they open, but on average a child will find it in the fourth box.
X geometric, E(X) = 3½. Find P(X > 6).
X ~ Geo(p), P(X ≤ 3) = 8191331. Find a) P(X > 3) b) P(1 < X ≤ 3).
5. Is a Binomial / Geometric Model Suitable?
The chapter's aim: recognise practical situations. Check the four conditions; if one fails, the model isn't suitable. Typical reasons it fails:
| Situation | Why NOT binomial / geometric |
|---|---|
| Height of the tallest of 3 people chosen from 10 | X is a measurement (not a count of successes) |
| Girls chosen when 2 children picked from 1 girl + 3 boys | no replacement ⇒ trials not independent, p changes |
| Motorbikes when 4 vehicles picked from car park (cars, buses, bicycles) | there are no motorbikes (p = 0) / no replacement, so p not constant |
| Monsoon months with > 5 m rain over 32 years | rainfall in months of one year is not independent, and p differs by month |
| Athlete's races until first win | p not constant (form, fitness, opponents vary); results not independent |
| Grain of rice dropped onto a chessboard until it lands on white | probably fine in theory, but it depends on where it is dropped — p may not be constant |
| Bag of 2 red + many green sweets; eat until 1st red | sweets are eaten (no replacement) so p changes; fine only if "many" ≈ replacement |
6. Explorations (Investigate!)
Frog starts bottom-left of a 5×5 grid, jumps only N or E to an adjacent lily pad. Number of routes to a pad: e.g. the pad with 3 routes → EEN, ENE, NEE. Sketch the 5×5 grid with the number of routes to all 24 pads; describe patterns; the flowered pads form a sequence — continue it and find the nth term.
Routes to a pad = routes to the pad below + routes to the pad on its left ⇒ Pascal's triangle along the diagonals; the grid (starting 1,1,1,1,1) is built that way.
| 1 | 5 | 15 | 35 | 70 |
| 1 | 4 | 10 | 20 | 35 |
| 1 | 3 | 6 | 10 | 15 |
| 1 | 2 | 3 | 4 | 5 |
| 🐸 | 1 | 1 | 1 | 1 |
Route counts (top row = north). Numbers on diagonals are the rows of Pascal's triangle; the number of routes to the pad a east and b north = (a+ba).
X = games you win in one game: P(0)=0.4, P(1)=0.6. Y = opponent's wins: P(0)=0.6, P(1)=0.4. First to win two games wins.
The stronger player gains the advantage as the number of games increases (0.6 → 0.648 → 0.683) — evidence: your probability of winning the contest rises with each longer contest.
7. Quick Recap — Whole Chapter on One Page
- Use for: number of successes in n repeated independent trials, p constant, two outcomes, n finite
- P(X = r) = (nr) pr (1−p)n−r, r = 0, 1, …, n [(nr) = n!r!(n−r)! = number of ways]
- E(X) = μ = np
- Var(X) = σ² = np(1−p) = npq, σ = √(npq)
- Find n, p from E & Var: q = Var ÷ E, p = 1 − q, n = E ÷ p
- "At least one" → 1 − P(X = 0) = 1 − qⁿ | unknown n → logs (flip sign when dividing by log of a number < 1)
- Use for: number of trials up to and including the first success; independent, p constant, can go on forever
- P(X = r) = p(1−p)r−1 = qr−1p, r = 1, 2, 3, … (r−1 failures, then a success)
- E(X) = μ = 1p
- P(X ≤ r) = 1 − qr P(X > r) = qr (q = 1 − p)
- Mode = 1 for all geometric distributions; P(X = r) decreases as r increases
- Probabilities form a GP (first term p, ratio q), and sum to 1
- "at least 6" = X > 5 | "fewer than 8" = X ≤ 7 | "at most 3" = X ≤ 3