Bernoulli trials and the binomial probability
Tossing a coin 5 times and counting heads, or checking 10 bulbs and counting defective ones, are repeated yes/no trials with the same chance each time.
A Bernoulli trial has two outcomes, success (probability p) and failure (q = 1 − p). In n independent trials the number of successes X has the binomial distribution B(n, p): P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, r = 0, 1, …, n.
Rule 1: Conditions — Fixed n, two outcomes, independent trials, constant p.
Rule 2: Formula — P(X = r) = ⁿCᵣ pʳ (1 − p)ⁿ⁻ʳ.
How is the formula obtained? : A fair coin is tossed 4 times: P(exactly 2 heads) = ⁴C₂ (½)² (½)² = 6/16 = 3/8.
- Easy : n = 4, p = ½: P(X = 2) = 3/8.
- Medium : A coin tossed 5 times: P(at least 4 heads) = ⁵C₄/32 + 1/32 = 6/32 = 3/16.
- Exam-type : A die thrown 6 times: P(exactly two sixes) = ⁶C₂(1/6)²(5/6)⁴ = 15 × 625/46656 ≈ 0.201.
Probabilities of all r from 0 to n add up to 1.
Forgetting ⁿCᵣ.
"At least" = 1 − P(none) when n is large. Memory clue : "n choose r, p to the r, q to the rest."
p and q must be for the same event across trials.
CET-style example : A fair coin is tossed 4 times; P(exactly 2 heads) is (A) 3/8 (B) 1/4 (C) 1/2 (D) 5/16
Answer : (A). Time-saving method : ⁴C₂ = 6, divide by 16.
Spot the difference
| Symbol | Meaning |
|---|---|
| n | number of trials |
| p | probability of success |
| q | 1 − p |
Practice questions
- 1Conditions of a binomial distribution?
- 2n = 3, p = ½: P(X = 0)?
- 3n = 4, p = ½: P(X = 2)?
- 4Coin tossed 5 times: P(exactly 3 heads)?
- 5Coin tossed 5 times: P(at least 4 heads)?
- 610 bulbs, each defective with p = 0.1: P(no defective)?
- 7Die thrown 6 times: P(exactly two sixes)?
- 8Coin tossed 3 times: P(at least one head)?
Show answers (try them yourself first)
- 1Fixed n, two outcomes, independence, constant p.
- 21/8.
- 33/8.
- 410/32 = 5/16.
- 53/16.
- 60.9¹⁰ ≈ 0.349.
- 7≈ 0.201.
- 81 − 1/8 = 7/8.
Mean, variance and standard deviation of the binomial distribution
If a basketball player makes 3 out of 5 shots on average, in 100 shots we expect about 60 baskets, with some natural spread around that.
For X ~ B(n, p): mean E(X) = np, variance Var(X) = npq, standard deviation σ = √(npq). Always 0 < q < 1 so Var(X) < E(X).
Rule 1: Mean — np.
Rule 2: Variance — npq.
How is the formula obtained? : n = 12, p = 1/3: mean = 12 × 1/3 = 4; variance = 12 × (1/3)(2/3) = 8/3; σ = √(8/3) ≈ 1.63.
- Easy : n = 10, p = 0.5: mean 5, variance 2.5.
- Medium : n = 12, p = 1/3: mean 4, variance 8/3.
- Exam-type : If mean = 6 and variance = 4.5 then q = 4.5/6 = 0.75, p = 0.25, n = 24.
Variance is smaller than the mean.
Using npq as the mean.
Find p from variance ÷ mean = q. Memory clue : "Mean np, spread npq."
If the stated variance exceeds the mean the data cannot be binomial.
CET-style example : For B(12, 1/3) the variance is (A) 8/3 (B) 4 (C) 3 (D) 2/3
Answer : (A). npq = 12 × 1/3 × 2/3. Time-saving method : Compute np then multiply by q.
Spot the difference
| Quantity | B(n, p) |
|---|---|
| Mean | np |
| Variance | npq |
| Standard deviation | √(npq) |
Practice questions
- 1Mean of B(10, 0.5)?
- 2Variance of B(10, 0.5)?
- 3Mean of B(12, 1/3)?
- 4Variance of B(12, 1/3)?
- 5σ of B(12, 1/3)?
- 6Mean 6 and variance 4.5: find p.
- 7Mean 6, variance 4.5: find n.
- 8Can a binomial have mean 4 and variance 5?
Show answers (try them yourself first)
- 15.
- 22.5.
- 34.
- 48/3.
- 5≈ 1.63.
- 6q = 0.75, p = 0.25.
- 7n = 6/0.25 = 24.
- 8No; variance npq < np.