A company has four electric generators. The time until failure for each generator follows an exponential distribution with mean 18 months. The company will begin using the other generator immediately after the current generator fails. Define random variable X as the total electricity time (in years) that the four generators can produce.
What is the probability that X is at least 6 years?
I understand that X follows gamma distribution, letting X~Gamma~(α = 4, λ = 1/1.5years = 2/3 year per generator)
Then we use possion distribution, letting W be the no. of generator needed to produce at least 6 years electricity time - [W~Poisson(μ = λ x 6 = 2/3 x 6 = 4)] to find P(X≥6) = P(1≤W≤4)
Could anyone let me know if my reasoning is logical to solve this question? Would be very grateful to receive any constructive feedback regarding my question. Thank you!
The probability that $X\geq6$ equals the probability that $X>6$ because $P(X=6)=0$.
Further it is more handy to work with $X>6$ in this context.
If $N(t)$ denotes a Poisson process equipped with rate $\lambda=12/18=2/3$ then the probability that you are looking for is: $$P(X>6)=P(X_1+X_2+X_3+X_4>6)=P(N(6)\leq3)=e^{-4}\sum_{k=0}^3\frac{4^k}{k!}$$ Here $N(6)$ corresponds with the rv $W$ in your question.
So the answer is not $P(1\leq W\leq 4)$ but is $P(W\leq 3)$.