find P( X$\lt$ Y) given X and Y are two independent exponential RV

The question is as follows:

You arrive at the post oﬃce and as you enter, each of the two clerks, Jim and Jack, starts serving a client. The amount of time needed by Jim to serve his client is X minutes, and the amount of time needed by Jack to serve his client is Y minutes, where X and Y are independent exponential random variables, X having parameter (rate) 1/2 per minute and Y having parameter 1/4 per minute.

Q) Since you are next in line, you will be served by whichever clerk is ﬁrst to ﬁnish serving his current client. Find the probability that you will be served by Jim.

So Jim is represented by X with rate = 1/2
Jack is represented by Y with rate 1/4

The question is asking for P ( X$$\lt Y)$$

a) I don't understand where the inner part of the integral $$P(X\lt Y | X=x)P(X=x)$$ is coming from.

b) How can you just replace P$$(x\lt Y)$$ with $$e^{\frac{-x}{4}}$$ ?

c) The p.d.f is $$\frac{1}{4}e^{\frac{-x}{4}}$$ where did the $$\frac{1}{4}$$ go?

The given solution are as follows:

b) If $$Y$$ is an exponential random variable with parameter $$\lambda$$, then $$P(Y>x) = e^{-\lambda x}$$. (If you remember the PDF, then you can re-derive this from scratch by $$P(Y>x) = \int_x^\infty \lambda e^{-\lambda t} \, dt = [e^{-\lambda t}]_{t=x}^\infty = e^{-\lambda x}$$.)
c) You ended up not needing the PDF of $$Y$$ in those computations.
• Yes $\mathsf P(X{<}Y)=\int_\Bbb R \mathsf p(X{<}Y, X{=}x)~\mathrm d x$. Then since the independence is stated in the problem, we have: $$\mathsf P(X{<}Y)~{=\int_\Bbb R \mathsf P(x{<}Y\mid X{=}x)~\mathsf p(X{=}x)~\mathrm d x\\=\int_\Bbb R \mathsf P(x{<}Y)~\mathsf p(X{=}x)~\mathrm d x}$$ Apr 22, 2019 at 11:53