I'm going through the MIT OCW probability course (6.041sc), but I'm having trouble on when to use CDF and the Poisson process. Here's the problem (Recitation 15, problem 1).
Problem Statement:
Beginning at time $t=0$, we begin using bulbs, one at a time, to illuminate a room. Bulbs are replaced immediately upon failure. Each new bulb is selected independently by an equally likely choice between a a type-A bulb and a type-B bulb. The lifetime, $X$, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF: \begin{aligned}\text{for type-A bulbs: }f_X(x) &= \begin{cases}e^{-x}, x\geq0,\\0, \text{ otherwise}\end{cases}\\\text{for type-B bulbs: }f_X(x) &= \begin{cases}3e^{-3x}, x\geq0,\\0, \text{ otherwise}\end{cases}\end{aligned}
Find the probability that there are no bulb failures before time $t$.
My Attempt:
I used the total probability theorem and then computed the CDF, $F_X(t)=P(X\leq t)$ : \begin{aligned}P(\text{no bulb failure before time }t)&=P(A)P(X\leq t|A)+P(B)P(X\leq t|B)\\&=\frac{1}{2}\int_0^t{e^{-x}}{dx}+\frac{1}{2}\int_0^t{3e^{-3x}}{dx}\\&=\frac{1}{2}\left(1-e^{-t}\right)+\frac{1}{2}\left(1-e^{-3t}\right)\end{aligned}
Solution:
\begin{aligned}P(\text{no bulb failure before time }t)=\frac{1}{2}e^{-t}+\frac{1}{2}e^{-3t}\end{aligned}
I was able to reproduce this result using the PMF for the number of arrivals $N_t$ in a Poisson process with rate $\lambda$, over an interval of length $t$. \begin{aligned}P_{N_t}(k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}, \text{ }k=0,1,\dots\end{aligned} In this context, we're looking at no arrivals, so $k=0$. And I figured that the arrival rate would be $\lambda=1,3$ for type-A(and type-B respectively) but I'm not sure why. Plugging in the appropriate numbers and using the total probability theorem we get the answer above.
My questions:
- Why did the CDF give me a different result? I'm sure that computing $P(X\leq t)$ was a valid approach, because that's the probability of the lifetime being at most $t$, but I must have some sort of conceptual misunderstanding on this.
- How would I know that the arrival rate for type-A is $1$(and $3$ for type-B)? The only way I'd think of that is the fact that both type A and B are exponentially distributed with parameter $\lambda=1,3$.
(1) Note your result is $$ \frac{1-e^{-t}}{2} + \frac{1-e^{-3t}}{2} = 1 - \frac{e^{-t}+e^{-3t}}{2} $$ since when you pick cdf, $F_X(t) = \mathbb{P}[X < t]$ is the probability bulb fails before $t$ and you need probability of no failures, so $1-F_X(t)$.
(2) Your argument makes sense.