Paper II · 4 Random Variables, Distributions & Inference · 4.1 Probability and probability distributions
4.1.2 Bernoulli, binomial and Poisson distributions
A Bernoulli trial has two outcomes; the binomial B(n, p) counts successes in n independent trials (mean np, variance npq), and the Poisson P(λ) counts rare events (mean = variance = λ) and is the limit of the binomial when n is large and p is small.
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Core facts
- Bernoulli trial: only two outcomes, success (probability p) or failure (probability q = 1 − p).
- Bernoulli distribution: p(x) = p^x q^(1−x), x = 0, 1. Mean p, variance pq. (James Bernoulli)
- Binomial distribution X ~ B(n, p): p(x) = ⁿCₓ p^x q^(n−x), x = 0, 1, …, n.
- Conditions: n fixed, trials independent, two outcomes each, p the same in every trial.
- Binomial mean = np, variance = npq, so variance < mean (q < 1).
- Binomial μ₃ = npq(q − p); β₁ = (q − p)²/(npq); β₂ = 3 + (1 − 6pq)/(npq).
- Symmetric when p = q = 1/2; positive skew if p < 1/2; negative skew if p > 1/2.
- Binomial mgf: (q + pe^t)^n.
- Poisson distribution X ~ P(λ): p(x) = e^(−λ) λ^x / x!, x = 0, 1, 2, … (S.D. Poisson, 1837).
- Poisson mean = variance = λ; μ₃ = λ; β₁ = 1/λ; β₂ = 3 + 1/λ.
- Poisson is always positively skewed and leptokurtic; it tends to symmetry as λ grows.
- Poisson mgf: e^(λ(e^t − 1)).
- Poisson as a limit of the binomial: n → ∞, p → 0, with np = λ fixed.
- Additive property: sum of independent B(n₁, p) and B(n₂, p) is B(n₁ + n₂, p) (same p); sum of independent P(λ₁) and P(λ₂) is P(λ₁ + λ₂).
Explained
1. Using the binomial formula. Find n and p, then count the ways (ⁿCₓ) and multiply.
Example 1. A fair coin is tossed 4 times. Find P(exactly 2 heads) and P(at least one head). Solution. n = 4, p = q = 1/2. P(X = 2) = ⁴C₂ (1/2)⁴ = 6/16 = 3/8. P(X ≥ 1) = 1 − P(X = 0) = 1 − 1/16. Answer: 3/8; 15/16
Example 2. A binomial variable has mean 4 and variance 3. Find n and p. Solution. q = npq/np = 3/4, so p = 1/4. n = 4/(1/4) = 16. Answer: n = 16, p = 1/4
A question that says "binomial with mean 3 and variance 4" has no answer: it would need q = 4/3 > 1.
Example 3. Three coins are tossed 80 times. Find the expected (theoretical) frequencies of 0, 1, 2, 3 heads. Solution. Expected frequency = N × ³Cₓ (1/2)³ = 80 × (1, 3, 3, 1)/8. Answer: 10, 30, 30, 10 This is how a binomial distribution is fitted to data: estimate p from the mean (p = x̄/n), then multiply N by each probability.
2. Using the Poisson formula. Poisson fits counts of rare events in a fixed time, space or lot: misprints per page, accidents per day, defects per item, telephone calls per minute, deaths by horse-kick (Bortkiewicz's classic data).
Example 4. Accidents at a junction follow a Poisson law with mean 2 per week. Find P(no accident) and P(at least one accident) in a week. Take e^(−2) = 0.1353. Solution. P(0) = e^(−2) = 0.1353. P(X ≥ 1) = 1 − 0.1353. Answer: 0.1353; 0.8647
Example 5. For a Poisson variable, P(X = 1) = P(X = 2). Find the mean. Solution. e^(−λ) λ = e^(−λ) λ²/2 gives λ = 2. Answer: 2 (In the same way, P(X = 2) = P(X = 3) gives λ = 3.)
Example 6. 2% of the bulbs made by a firm are defective. Use the Poisson approximation to find the chance that a box of 100 has no defective bulb. Solution. λ = np = 100 × 0.02 = 2. P(0) = e^(−2) ≈ 0.135. (The exact binomial value 0.98¹⁰⁰ ≈ 0.133 is very close.) Answer: about 0.135
Deeper
1. Comparison.
| Point | Bernoulli | Binomial B(n, p) | Poisson P(λ) |
|---|---|---|---|
| Values | 0, 1 | 0, 1, …, n | 0, 1, 2, … (no upper limit) |
| Mean | p | np | λ |
| Variance | pq | npq (< mean) | λ (= mean) |
| Parameters | p | n, p | λ only |
Mean > variance suggests binomial. Mean = variance suggests Poisson. (Mean < variance points to the negative binomial, not covered here.)
2. Mean and variance of the binomial (sketch). X is the sum of n independent Bernoulli variables. Each has mean p and variance pq. By the addition theorem E(X) = np. Because the trials are independent, the variances add: V(X) = npq. The largest variance is n/4, at p = 1/2.
3. Recurrence relations (used to build tables quickly).
- Binomial: p(x + 1) = [(n − x)/(x + 1)] (p/q) p(x).
- Poisson: p(x + 1) = [λ/(x + 1)] p(x).
Start from p(0) = qⁿ or e^(−λ), then step upward.
4. Mode.
- Binomial: look at (n + 1)p.
- Not an integer: mode = its integral part. B(10, 1/2): (n + 1)p = 5.5, so the mode is 5.
- An integer m: bimodal, modes at m − 1 and m. B(5, 1/3): (n + 1)p = 2, modes 1 and 2 (both have probability 80/243).
- Poisson: λ not an integer: mode = integral part of λ. λ an integer: bimodal at λ − 1 and λ.
5. Derivation of the Poisson limit (sketch). Put p = λ/n in ⁿCₓ p^x q^(n−x). As n → ∞, n(n − 1)…(n − x + 1)/nˣ → 1, (1 − λ/n)^(−x) → 1, and (1 − λ/n)ⁿ → e^(−λ). What is left is e^(−λ) λˣ/x!.
6. Moments of the Poisson (sketch). Σ x e^(−λ) λˣ/x! = λ e^(−λ) Σ λ^(x−1)/(x − 1)! = λ e^(−λ) e^λ = λ. In the same way E[X(X − 1)] = λ², so E(X²) = λ² + λ and V(X) = λ.
7. Limits to the normal. The binomial tends to the normal when n is large and p is not close to 0 or 1. The Poisson tends to the normal as λ → ∞ (since β₁ = 1/λ → 0 and β₂ → 3). See STA-PD-04.
Example 7. Six fair coins are tossed. Find P(at least 5 heads). Solution. (⁶C₅ + ⁶C₆)/2⁶ = (6 + 1)/64. Answer: 7/64
Common traps
- ✗ "Binomial variance = np." ✓ Variance = npq; mean = np.
- ✗ "A binomial can have mean 3 and variance 4." ✓ Impossible: binomial variance is always less than the mean.
- ✗ "Poisson mean = λ, variance = √λ." ✓ Variance = λ; SD = √λ.
- ✗ "The binomial is always symmetric." ✓ Only when p = 1/2.
- ✗ "The Poisson can be negatively skewed." ✓ β₁ = 1/λ with μ₃ = λ > 0: always positively skewed.
- ✗ "Poisson limit: n → ∞ and p → 1." ✓ p → 0, with np = λ finite.
- ✗ "Any two binomials add to a binomial." ✓ Only when they have the same p (and are independent).
- ✗ "If (n + 1)p is an integer m, the mode is m." ✓ The distribution is bimodal: m − 1 and m.
- ✗ "In a binomial the trials may have different p." ✓ p must be constant across trials.
Quick revision
- Bernoulli: mean p, variance pq.
- B(n, p): ⁿCₓ pˣ qⁿ⁻ˣ; mean np; variance npq < mean.
- Binomial symmetric only at p = 1/2; max variance n/4.
- β₁ (binomial) = (q − p)²/npq; β₂ = 3 + (1 − 6pq)/npq.
- P(λ): e^(−λ)λˣ/x!; mean = variance = λ.
- Poisson: β₁ = 1/λ, β₂ = 3 + 1/λ; always positive skew.
- Poisson = binomial with n → ∞, p → 0, np = λ.
- Mode: integral part of (n + 1)p or λ; bimodal if integer.
- Recurrence: binomial (n − x)/(x + 1) · p/q; Poisson λ/(x + 1).
- Additive: B(n₁ + n₂, p); P(λ₁ + λ₂).
Practise this topic. Questions on Bernoulli & binomial distributions, with answers explained.
All Probability and probability distributions notes · TGPSC MPSO / ASO syllabus · TGPSC ASO Paper I Mock 1 with answers · Start free
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