For a binomial variable X with n trials and success probability p, P(X = r) = nCr × p^r × q^(n − r), where q = 1 − p. The formula multiplies the number of ways to choose which trials succeed by the probability of one such pattern.
This lesson follows recognising binomial conditions in the probability distributions chapter. The nCr part comes from permutations and combinations.
How do I set up the formula?
Write X ~ B(n, p) first, and name n, p and q. Then write the formula with the values in place, before using a calculator. This shows the method marks and catches swapped values early.
Worked example: a multiple-choice quiz
A quiz has 6 questions, each with 4 choices and exactly one correct. A student guesses every answer. Let X be the number correct, so X ~ B(6, 0.25) and q = 0.75.
Exactly 2 correct.
P(X = 2) = 6C2 × 0.25² × 0.75⁴ = 15 × 0.0625 × 0.31640625 = 0.2966 (4 d.p.).
At least 1 correct. Use the complement:
P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.75⁶ = 1 − 0.17798 = 0.8220.
At least 4 correct. Add the three cases.
- P(X = 4) = 6C4 × 0.25⁴ × 0.75² = 15 × 0.00390625 × 0.5625 = 0.03296
- P(X = 5) = 6C5 × 0.25⁵ × 0.75 = 6 × 0.0009765625 × 0.75 = 0.00439
- P(X = 6) = 0.25⁶ = 0.00024
The total is 0.03296 + 0.00439 + 0.00024 = 0.0376 (4 d.p.).
The mistake that costs marks
The common slip is to swap p and q, or to leave out nCr. Both give an answer that still looks like a probability, so the error is easy to miss.
| Part | Wrong | Right |
|---|---|---|
| Choosing the trials | left out | 6C2 = 15 |
| Success factor | 0.75² | 0.25² |
| Failure factor | 0.25⁴ | 0.75⁴ |
| Result | small, unchecked | 0.2966 |
A quick check: the most likely count should sit near np = 1.5. So P(X = 2) being about 0.30 is plausible, and a result near 0.002 is not.
Check yourself
A factory finds that 10% of bulbs are faulty. A box holds 5 bulbs chosen independently. Find the probability that exactly 1 is faulty, and the probability that none is faulty.
Answer
X ~ B(5, 0.1) and q = 0.9.
P(X = 1) = 5C1 × 0.1 × 0.9⁴ = 5 × 0.1 × 0.6561 = 0.3281 (4 d.p.).
P(X = 0) = 0.9⁵ = 0.5905 (4 d.p.).
Check: these two plus the remaining cases must total 1, and 0.3281 + 0.5905 = 0.9186 leaves 0.0814 for 2 or more faulty bulbs, which is plausible.
What to study next
Continue with finding mean and variance of a binomial variable, which summarises the whole distribution in two numbers. The word-problem structure worksheet helps you find n, p and the event in a long question.
If setting up binomial questions is the hard step, see online one-to-one Additional Mathematics tuition.