Question

Two faulty machines, M1 and M2, are repeatedly run synchronously in parallel (i.e., both machines execute one run, then both execute a second run, and so on). On each run, M1 fails with probability p1 and M2 fails with probability p2, all failure events being independent. Let the random variables X1, X2 denote the number of runs until the first failure of M1, M2 respectively; thus X1, X2 have geometric distributions with parameters p1, p2 respectively. Let X denote the number of runs until the first failure of either machine. Compute the distribution of X. What is its expectation?

Answer 1

The event that either M1 or M2 fails has probability

`P(M_1\text{ fails or }M_2\text{ fails})=P(M_1\text{ fails})+P(M_2\text{ fails})-P(M_1\text{ and }M_2\text{ both fail})`

by the addition rule. Failure events are independent, so

`P(M_1\text{ and }M_2\text{ both fail})=P(M_1\text{ fails})P(M_2\text{ fails})`

so that

`P(M_1\text{ fails or }M_2\text{ fails})=p_1+p_2-p_1p_2`

Denote this probability by
`p`. Then
`X` follows a geometric distribution with this parameter
`p` and has density

`P(X=x)=\begin{cases}(1-p)^(x-1)p&\text{for }x\ge1\\0&\text{otherwise}\end{cases}`

The expectation is
`\frac1p=\frac1{p_1+p_2-p_1p_2}`.