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Assume that we have a serial circuit with three bulbs. Each bulb's life time is exponentially distributed: $$f_{bulb}(t) =\left\{ \begin{aligned} &\lambda e^{-\lambda t} & t \ge 0\\ &0 & t < 0 \end{aligned} \right.$$ Find the distribution function of the circuit's work-time (the circuit works when all three bulbs work). \begin{align} t &= \min \{t_1,t_2,t_3\}\\ F_{circuit}(t) &= P(T_1<t, T_2 < t,T_3 < t) \end{align}

I'm stuck here. Did I do this correctly? If yes how do i proceed?

(I don't know how to process the $\min\{t_1,t_2,t_3\}$ part)

Sorry if i have bad grammar or messy notation.

Hùng Nguyễn
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  • Hint: If you have three independent events $A$, $B$, $C$, then $P(A$ and $B$ and $C)=P(A)\cdot P(B) \cdot P(C)$. Let $A$ be the event that $T_1<t$, let $B$ be the event that $T_2<t$, and let $C$ be the event that $T_3<t$. – irchans Apr 18 '19 at 08:52
  • Sorry, that wasn't the part that bugging me, i edited the problem to be more specific. – Hùng Nguyễn Apr 18 '19 at 09:00
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    This is relevant: https://math.stackexchange.com/questions/580279/how-to-prove-that-the-minimum-of-two-exponential-random-variables-is-another – saulspatz Apr 18 '19 at 09:07
  • @saulspatz That was extremely helpful. Thank you. – Hùng Nguyễn Apr 18 '19 at 09:38

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