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Suppose we have a system like the one in the picture. Each component works and fails independently, and each component's duration is given by a random variable with exponential distribution with $\lambda = 0.002$.

What's the probability that the system will last more than $500$ hours?

enter image description here

I tried the following, but I'm not sure if its correct

Let $T_i$ be the duration of each component. We know that the system fails if component $1$ fails, or if both, components $2$ and $3$ fail (the two in parallel).

Therefore $P(T<t)= P(T_1<t) + P(T_2,T_3<t)- P(T_1,T_2,T_3<t)$, this yields (through independence), $P(T<t)=(1-\exp(-\lambda t))+\left(1-\exp(-\lambda t)\right)^2-(1-\exp(-\lambda t))^3$.

And now, I just do $P(T>500)=1-P(T<500)\approx 0.22$.

Is this correct?

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Let $T_i$ be the lifetime of component $i$. Then the lifetime of the system is $$T = T_1\wedge(T_2\vee T_3)$$ The $T_i$ are independent with distribution function $F(t)=1-e^{-\lambda t}$, so $T_2\vee T_3$ has distribution function $F^2(t)=\left(1-e^{-\lambda t}\right)^2$. It follows then that $T$ has survivor function \begin{align} \overline G(t) &=(1-F(t))(1-F^2(t))\\ &= e^{-\lambda t}\left(1-\left(1-e^{-\lambda t}\right)\right)\\ &= 2e^{-2\lambda t} - e^{-3\lambda t} \end{align} A straightforward computation yields $$\mathbb P(T>500) = \overline G(500) = 2e^{-2}-e^{-3}\approx 0.220883.$$

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Comment: Verification, by simulation (2 or3-place accuracy).

m = 10^6
a = rexp(m, .002)
b = rexp(m, .002)
c = rexp(m, .002)
mean((a > 500) & (b > 500 | c > 500))     ## '&' means 'and';  '|' means 'or'
## 0.221315
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