# exponential random variable

A system has $3$ components and the system works if at least $2$ of $3$ components work. Life time of components are i.i.d. exponential random variable with mean $1$. If $X$ denotes life time of the system, then $E(X)$ is

a. $1$

b. $2/3$

c. $5/6$

d. $1/2$

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You may want to start by thinking about the distribution of $X$, the life time of the system. The system will have failed by time $t$ if and only if either two or all three components have failed by that time. Hence

$$P(X\leq t) = P(\text{Two or three components have failed by time t}) = P(\text{Two components have failed by time t) + P(Three components have failed by time t}).$$ For the first probability you may first think about the situation that component one and two failed by time $t$, while component three is still working. Because life times are independent this gives $$P(\text{Comp 1 and 2 failed by time t, comp 3 still alive}) = [1-\exp(-t)]^2\exp(-t).$$ Considering also the cases that components one or two may be the ones still alive by time $t$, you find that $$P(\text{Two components have failed by time t}) = 3[1-\exp(-t)]^2\exp(-t).$$ Similarly one obtains $$P(\text{Three components have failed by time t}) = [1-\exp(-t)]^3.$$

This should allow you to find $P(X\leq t)$ and then $EX$.

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Thanks, I edited my answer. –  Eckhard Dec 17 '12 at 14:04
Let $X_{i}$ denote life time of i's component, and $X$ denote life time of the system as your notations. $X$ will work if at least two components of it work. In other words It fails to work if two components will fail to work. This means at time the life of system finished second component is failed to work (pay attention I'm not pointing to $X_{2}$, the mean is second component which is stopped). So if life time of $X$ be equal to $x$ means before time $x$ one of components is stopped, at time $x$ one different component is stopped, and the last component is stopped after time $x$. Therefore $X=x$ will accurse if $$(x_{1},x_{2},x_{3})\in \{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}=x,x_{2}\leq x,x_{3}\geq x\}\cup \{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}=x,x_{2}\geq x,x_{3}\leq x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\leq x,x_{2}=x,x_{3}\geq x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\geq x,x_{2}=x,x_{3}\leq x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\leq x,x_{2}\geq x,x_{3}=x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\geq x,x_{2}\leq x,x_{3}=x\}$$ Because Life time of components are i.i.d. exponential random variables with mean 1, so $$f_{X}(x)=e^{-x}\int_{0}^{x}e^{-x_{2}}dx_{2}\int_{x}^{\infty}e^{-x_{3}}dx_{3}+e^{-x}\int_{x}^{\infty}e^{-x_{2}}dx_{2}\int_{0}^{x}e^{-x_{3}}dx_{3}+(\int_{0}^{x}e^{-x_{1}}dx_{1})e^{-x}(\int_{x}^{\infty}e^{-x_{3}}dx_{3})+(\int_{x}^{\infty}e^{-x_{1}}dx_{1})e^{-x}(\int_{0}^{x}e^{-x_{3}}dx_{3})+(\int_{0}^{x}e^{-x_{1}}dx_{1}\int_{x}^{\infty}e^{-x_{2}}dx_{2})e^{-x}+(\int_{x}^{\infty}e^{-x_{1}}dx_{1}\int_{0}^{x}e^{-x_{2}}dx_{2})e^{-x}$$ By calculating above integrals we have $$\forall x\in[0,\infty)\; :\; f_{X}(x)=6(e^{-2x}-e^{-3x})$$ So $$E(X)=\int_{0}^{\infty}6(e^{-2x}-e^{-3x})xdx=6\int_{0}^{\infty}xe^{-2x}dx-6\int_{0}^{\infty}xe^{-3x}dx=6(\frac{1}{4})-6(\frac{1}{9})=\frac{3}{2}-\frac{2}{3}=\frac{5}{6}$$ So the choice "c" is correct.
-1 for giving the full answer and because $P(X=x)=0$ for every $x$. –  Did Jan 8 '13 at 9:34