Independent events
Definition taught in Random experiments and probability, part of Probability
Independent events
| Type | Definition |
| Taught in | Random experiments and probability |
| Part | Probability |
| Properties | 1 |
| Worked examples | 2 |
Events A_1,A_2,\ldots are mutually independent when, for every finite subcollection,
\mathbb{P}(A_{i_1}\cap\cdots\cap A_{i_k})=\mathbb{P}(A_{i_1})\cdots\mathbb{P}(A_{i_k}).
Properties
Product rule
Main article: Product rule
If \mathbb{P}(A_1\cap\cdots\cap A_{n-1})>0, then
\mathbb{P}(A_1\cap\cdots\cap A_n) =\mathbb{P}(A_1)\,\mathbb{P}(A_2\mid A_1)\,\mathbb{P}(A_3\mid A_1\cap A_2)\cdots\mathbb{P}(A_n\mid A_1\cap\cdots\cap A_{n-1}).
Each factor is the definition of conditional probability, and the intermediate intersections cancel.
Examples
A four-engine aeroplane
Worked example from Random experiments and probability
An aeroplane has engines 1 and 2 on the left wing and 3 and 4 on the right. It can fly as long as at least one engine on each wing works. Each engine fails during a flight with probability q, independently of the others.
Question. Describe the sample space and the event that the aeroplane can fly, and find its probability.
Solution. Record which engines work: an outcome is a vector in \{0,1\}^4, so there are 2^4=16 outcomes. Let A_i be the event that engine i works. The aeroplane can fly on the event
(A_1\cup A_2)\cap(A_3\cup A_4).
A wing fails only if both its engines fail, with probability q^2, so by independence of the two wings
\mathbb{P}\{(A_1\cup A_2)\cap(A_3\cup A_4)\}=(1-q^2)^2.
With q=0.1 that is 0.9801. A single engine fails one flight in ten, but the aeroplane only about one flight in fifty. Redundancy turns unreliable parts into a reliable system, which is the subject of Reliability.
Three components, one of them essential
Question page: Three components, one of them essential
Components 1, 2 and 3 fail independently with probability 0.1 each. The system works when component 1 works and at least one of components 2 and 3 works. Find the probability that the system works.
Let A_i be the event that component i works, so \mathbb{P}(A_i)=0.9. The system works on A_1\cap(A_2\cup A_3). Independence gives
\mathbb{P}(A_2\cup A_3)=0.9+0.9-0.9^2=0.99,
and the system probability is 0.9\times 0.99=0.891.
Used in
- Taught in Random experiments and probability, where the derivation and examples are.
- All 17 questions on Random experiments and probability
See also
Sample space · Event · Probability measure · Properties of a probability · Equally likely outcomes · Four ways of counting draws · Conditional probability · Product rule