Probability calculator

The rules

• Complement: P(not A) = 1 − P(A).
• Addition rule: P(A or B) = P(A) + P(B) − P(A and B). The joint probability is subtracted because it would otherwise be counted twice. For mutually exclusive events, which cannot happen together, P(A and B) = 0 and the rule becomes P(A) + P(B).
• Multiplication rule: P(A and B) = P(A) × P(B given A). For independent events, where one does not change the chance of the other, P(B given A) = P(B), so P(A and B) = P(A) × P(B).
• Conditional probability: P(A given B) = P(A and B) / P(B).

Mutually exclusive and independent are often confused. They are nearly opposites: if two events with nonzero probability are mutually exclusive, knowing one happened tells you the other did not, so they cannot be independent.

The worked example

Sixty percent of parts come from line 1. Five percent of line 1 parts are defective. So P(line 1 and defective) = 0.60 × 0.05 = 0.03. If four percent of all parts are defective, then P(line 1 given defective) = 0.03 / 0.04 = 0.75. Three quarters of the defectives come from line 1, which makes 60% of the parts. The calculator also shows that the events are not independent.

Counting

• Combinations, where order does not matter: nCr = n! / (r!(n − r)!). Ten parts, choose three for test: 120 ways.
• Permutations, where order matters: nPr = n! / (n − r)!. Ten candidates for first, second and third place: 720 ways.

At least one

The chance of at least one occurrence in n independent tries is 1 − (1 − p)ⁿ, the complement of "none in n". With a 2% chance each time, the chance of at least one in 50 tries is 64%. It is not 100%, and it is not 50 × 2%.

Series and parallel reliability

• Series: the system works only if every part works. R_s = R₁ × R₂ × …. A series system is less reliable than its weakest part.
• Parallel (redundant): the system works if any part works. R_p = 1 − (1 − R₁)(1 − R₂)….

Mixed systems are reduced block by block: work out each parallel block, then multiply the blocks in series. Redundancy only helps as much as the failures are independent. The fault tree shows the same logic from the failure side.

On the exam

CQPA asks you to use independent and mutually exclusive events, combinations, permutations, the addition and multiplication rules and conditional probability (III.A.3, Apply). CQE (VI.B.3) and CSSBB (V.E.1) cover the same concepts, and CQE includes reliability of systems (III.E). Expect at least one question that turns on whether events are independent or mutually exclusive.

Common questions

How do I calculate the probability of at least one?

For independent trials, P(at least one) = 1 − (1 − p)ⁿ, because the only way to get none is to miss every time. With a 2% chance per part and 10 parts, the chance of at least one is 1 − 0.98¹⁰, about 0.183.

What is the difference between combinations and permutations?

Combinations count selections where order does not matter: nCr = n! / (r!(n − r)!). Permutations count arrangements where order matters: nPr = n! / (n − r)!. Choosing 3 parts from 10 for a sample is a combination; assigning 3 of 10 parts to three different tests in order is a permutation.

How do I calculate series and parallel reliability?

Components in series all have to work, so multiply their reliabilities: two units at 0.9 give 0.81. Redundant components in parallel fail only if all fail, so the reliability is 1 minus the product of the unreliabilities: two units at 0.9 give 1 − 0.1 × 0.1 = 0.99.

Where is the probability distribution calculator?

This page works the probability rules. For binomial, Poisson, hypergeometric, normal, Weibull and other distributions, use the Probability distribution calculator, which gives the probabilities and draws the curve.

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