Probability, Combinations and Binomial Formulas
Probability formulas for reliability and test work: two events, repeated trials, permutations, combinations and the binomial distribution, with worked examples.
The formulas
| To find | Formula |
|---|---|
| Both of two independent events | P(A and B) = P(A) × P(B) |
| Either of two events | P(A or B) = P(A) + P(B) − P(A and B) |
| Neither event | P(neither) = 1 − P(A or B) |
| Conditional probability | P(A given B) = P(A and B) / P(B) |
| At least once in n independent trials | 1 − (1 − p)ⁿ |
| Permutations of r from n (order matters) | nPr = n! / (n − r)! |
| Combinations of r from n (order does not matter) | nCr = n! / (r! (n − r)!) |
| Exactly k successes in n trials | P(X = k) = nCk pᵏ (1 − p)ⁿ⁻ᵏ |
| Mean and variance of the count | n p and n p (1 − p) |
| At least k successes | P(X ≥ k) = sum of P(X = j) for j = k to n |
Every formula here assumes independence: the outcome of one trial does not change the chance of another. Two lamps on one circuit, or two units fed from one supply, can fail together, and the independent answer is then too optimistic. Multiplying the chances of a series of events gives the chance that all of them happen; for a redundant pair that both must fail, the product is what you want. A small probability over many trials still adds up, which is why the at-least-once formula matters more than the single-trial figure. Probabilities are numbers from 0 to 1; multiply by 100 for a percentage.
Worked example: two redundant units
Two independent units each have a 0.02 chance of failing in a year.
- Both fail: 0.02 × 0.02 = 0.0004
- At least one fails: 0.02 + 0.02 − 0.0004 = 0.0396
- A single unit failing at least once in 5 years: 1 − (1 − 0.02)^5 = 0.09608
Worked example: counting selections
How many ways can three of ten spare units be picked and put in order, and without order?
- In order: 10P3 = 10 × 9 × 8 = 720
- Without order: 10C3 = 720 / 3! = 120
- Each group of three can be arranged in 3! = 6 orders, which is why the first figure is six times the second
Worked example: lamp failures
A row of 20 lamps, each with a 0.05 chance of failing in a year, treated as independent.
- Exactly 2 fail: 20C2 × 0.05² × 0.95¹⁸ = 190 × 0.0025 × 0.3972 = 0.1887
- At most 2 fail: P(0) + P(1) + P(2) = 0.3585 + 0.3774 + 0.1887 = 0.9245
- At least 3 fail: 1 − 0.9245 = 0.07548
- Mean number failing = n p = 1 and variance = n p (1 − p) = 0.95
Frequently asked questions
What is the difference between a permutation and a combination?
A permutation counts different orders as different selections; a combination counts them as one. Choosing three units to place in three positions is a permutation; choosing three units to keep as spares is a combination.
When can I multiply probabilities?
When the events are independent. If one event makes the other more or less likely, as with two units on a shared supply, use the conditional probability P(A given B) in the product instead.
Why use the binomial distribution?
It gives the chance of exactly k successes in n independent trials with the same chance each time. For a k-out-of-n redundant system, count a working unit as a success and read the at-least-k figure.
Does a low failure probability mean failures will not happen?
No. With many units or many years the chance of at least one failure grows, as 1 − (1 − p)ⁿ shows. Twenty lamps at 5% each give a mean of one failure a year.
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