Two outcomes, repeated
A Bernoulli trial has exactly two outcomes: success or failure. A binomial experiment repeats such trials.
Conditions (BINS):
- Binary: two outcomes each trial
- Independent trials
- Number of trials fixed
- Same probability of success p each time
The formula
P(X = k) = nCk × pᵏ × (1 − p)ⁿ⁻ᵏ
- n = number of trials
- k = number of successes
- p = probability of success
- nCk counts the ways to arrange the successes
Example
A basketball player makes 70% of free throws. In 5 shots, what’s P(exactly 3 made)?
- n = 5, k = 3, p = 0.7
- 5C3 = 10
- P = 10 × 0.7³ × 0.3² = 10 × 0.343 × 0.09 = 0.3087
Why nCk?
Making 3 of 5 could happen as MMMXX, MXMXM, XXMMM, and so on. Each has probability 0.7³ × 0.3², and there are 5C3 = 10 such orders.
"At least" and "at most"
At least 1 success: 1 − P(0)
Example: P(at least 1 six in 4 rolls) = 1 − (5/6)⁴ ≈ 0.518
At most 2: P(0) + P(1) + P(2)
The binomial distribution
Listing P(X = k) for all k gives the binomial distribution.
Example: n = 4 coin flips, p = 0.5
| Heads k | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P | 1/16 | 4/16 | 6/16 | 4/16 | 1/16 |
The numerators 1, 4, 6, 4, 1 come from Pascal’s triangle.
Mean and spread
- Expected value (mean): μ = np
- Standard deviation: σ = √[np(1 − p)]
Example: 100 coin flips → μ = 50 heads, σ = 5.
Technology
Spreadsheets: =BINOM.DIST(k, n, p, FALSE) for exactly k; TRUE for at most k. Calculators have binompdf and binomcdf.
Not binomial
- Drawing cards without replacement (p changes)
- Counting until the first success (n isn’t fixed)
Worked examples
Example 1: P(exactly 2 heads in 3 flips)? 3C2 × 0.5³ = 3/8.
Example 2: A test has 10 multiple-choice questions with 4 options each. Guessing, P(exactly 5 correct)? 10C5 × 0.25⁵ × 0.75⁵ ≈ 0.058.
Example 3: 20% of bulbs are defective. Expected defectives in 50? 10.
Common mistakes
- Forgetting nCk.
- Using the formula when trials aren’t independent.
- Mixing up "at least" and "exactly".
Try it yourself
- P(exactly 4 heads in 6 flips)?
- p = 0.2, n = 5. P(0 successes)?
- P(at least 1 success) for question 2?
- n = 200, p = 0.3. Mean and standard deviation?
Answers: 1) 15/64 ≈ 0.234 2) 0.8⁵ ≈ 0.328 3) ≈ 0.672 4) 60 and ≈ 6.48
Quick recap
- Binomial probability: P(X = k) = nCk pᵏ (1 − p)ⁿ⁻ᵏ.
- Requires fixed, independent Bernoulli trials with constant p.
- Mean (expected value) np; SD √[np(1 − p)].