$X$ and $Y$ are exponential variates. How do I find $P(X<Y)$

Question

The lifetime of two brands of bulbs $X$ and $Y$ are exponentially distributed with the mean life of $100$ hours. Bulb $X$ is switched on $15$ hours after Bulb $Y$ has been switched on. The probability that bulb $X$ fails before Bulb $Y$ is?

The options are:

$(A). \frac{15}{100}$

$(B). \frac{1}{2}$

$(C).\frac{85}{100}$

$(D). 0$

Answer 1

Edit: (Bulb X is switched on 15 hours after bulb Y)

You must partition on the 15 hour.   Bulb-X may fail before bulb-Y, only if bulb-Y does not fail before bulb X is switched on. $$\mathsf P(X<Y)~=~\mathsf P(Y\geq 15)\cdot\mathsf P(X<Y\mid Y\geq 15)$$ Now, use the memoryless property to simplify.

Answer 2

If bulb $Y$ might fail before bulb $X$ is switched on,

\begin{align} &\text{ probability that bulb $X$ fails before bulb } Y \\ &= P(Y \geq 15) P(X<Y-15|Y \geq 15)+P(Y < 15)(0) \\&=P(Y \geq 15) P(X < Y) \\ &= \frac12 P(Y \geq 15) \end{align}

If we are given that bulb $Y$ is still working when bulb $X$ is switched on \begin{align} &\text{ probability that bulb $X$ fails before bulb } Y \text{ given that bulb } Y \text{ was functioning when bulb } X \text{was switched on} \\ &= P(X<Y-15|Y \geq 15) \\&= P(X < Y) \\&= \frac12 \end{align}