Two exponentially distributed random variables w/ different intensity. Which is more probable to take?

Question

Let's say I have two types of light bulbs, A which has $E(A)=100$ hours of lifetime, and B which has $E(B)=200$. I have three of type A and one of type B. I randomly use one of the four, and after 200 it still works. What is the probability that I chose one of type A?

So, $A$~Exp(0.01) and $B$~Exp(0.005), and $P(X≤t)=F(t)=1-e^{-ßt}$ for $X$~Exp(ß). What's the first step here? Is the probability the same for any time interval as exponential distribution is memoryless?

Answer 1

First, we note that $P(A \ge t) = e^{-t/100}$ and $P(B \ge t) = e^{-t/200}$. Next, denote $X$ as the chosen lightbulb. We have \begin{align*} P(X = A|X \ge 200) &= \frac{P(X \ge 200|X = A) P(X = A)}{P(X \ge 200|X = A)P(X = A)+P(X \ge 200|X = B)P(X = B)} \\ &=\frac{e^{-200/100}\cdot \frac{3}{4}}{e^{-200/100}\cdot\frac{3}{4}+e^{-200/200}\cdot \frac{1}{4}} \\ &\approx 0.525 \end{align*}