A computer manufacturer produces two types of laptops, Machine A and Machine B. They offer an automatic manufacturer's warranty of 3 years when a consumer buys one of their laptops, good for two free repairs and a brand new replacement if the purchased laptop breaks down a third time (within the first 3 years after purchase). For Machine A, the breakdowns occur according to a Poisson process with one breakdown every 2 years, on average. For Machine B, the breakdowns occur according to a Poisson process with one breakdown every 3 years, on average. Suppose that 70% of the laptops produced by the manufacturer are of Machine A type, while the rest are of Machine B type.
(a) Given that a particular laptop has functioned without need for repair for 2.5 years, what is the probability that the laptop is of Machine A type?
(b) What is the probability that a random laptop must be replaced for free under the rules of the manufacturer's warranty?
For a), I did $$P(A|Z)= \dfrac{P(Z|A)P(A)}{P(Z|A)P(A)+P(Z|B)P(B)}$$ where Z = a particular laptop that has functioned without need for repair for 2.5 years, A = Machine A type and B = Machine B type.
So P(A) = 0.7 and P(B) = 0.3. I used the Poisson distribution to find P(Z|A) and P(Z|B). For P(Z|A), lambda is 0.5 and l is 2.5 evaluated at x=0. For P(Z|B), lambda is $\frac{1}{3}$ and l is 2.5 evaluated at x=0. So I got P(Z|A) = $e^{-1.25}$ and P(Z|B) = $e^{\frac{5}{6}}$. So subbing everything back in, I got 0.606024.
$$ $$ For part b), I did the poisson distribution again but using different ls.
I used the equation $$P(Y)= 0.7P(X_{A}>2)+0.3P(X_{B}>2)$$ where Y = a random laptop will be replaced for free and $X_{A}$ and $X_{B}$ are the poisson distributions.
So P($X_{A}$>2) = 1 - P(X=0)- P(X=1)- P(X=2) = 1- $e^{-1.5} - 1.5e^{-1.5} - \frac{1.5^{2}e^{-1.5}}{2!}$
and P($X_{B}$>2) = 1 - P(X=0)- P(X=1)- P(X=2) = 1- $e^{-1}-e^{-1}-\frac{e^{-1}}{2!}$
So even though I solved it, I'm not sure if my answer is right. I'm also thinking if this is Gamma distribution or not. Can someone check over my work? Thanks.