Determine "winner" in exponential contest If I have one light bulb that could be one of $2$ kinds, ($A$ and $B$ are the lifetimes of first and second type: $A\sim \exp(1)$ and $B\sim \exp(3)$), and each time a bulb dies, another bulb replaces it (with probability $0.5$ to be $A$ or $B$). $X$ is the lifetime of the light bulb (not knowing which type it is). The initial light bulb could be $A$ or $B$ with probability $0.5$. Knowing that the bulb didn't die until time $t$, what is the probability that the light bulb type is the first type (with lifetime $A$)?

You have $\pi(A)=\pi(B)=0.5$. You want the posterior value $$\pi(A\mid\text{failure at }t)=\dfrac{\pi(A)f_A(t)}{\pi(A)f_A(t)+\pi(B)f_B(t)}.$$

The densities are $f_A=e^{-t}$ and $f_B=3e^{-3t}$ so $$\Pr(\text{type }A\mid\text{failure at }t) = \dfrac{0.5 \times e^{-t}}{0.5 \times e^{-t}+0.5 \times 3e^{-3t}}= \dfrac{1}{1+ 3e^{-2t}}.$$