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}, xgeq0,\0, text{ otherwise}end{cases}\text{for type-B bulbs: }f_X(x) &= begin{cases}3e^{-3x}, xgeq0,\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**, $F_X(t)=P(Xleq t)$

:

begin{aligned}P(text{no bulb failure before time }t)&=P(A)P(Xleq t|A)+P(B)P(Xleq 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,dotsend{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 $P(Xleq t)$ 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 $lambda=1,3$.