I am having an issue grasping the concepts on how to apply independent experiments to this problem. I know that our set that we look at is $50 \leq x \leq 60$ but not sure how to approach this other than that. Since they are independent we can split up the and statements $P(AB) = P(A)P(B)$ is it like $P(e^{−0.003})P(e^{−0.005x})$ but how would I plug in the hours?
Transistors are purchased from two independent sources $A$ and $B$. The probability that one purchased from $A$ will not fail prior to $x$ hours of operation is $e^{−0.003}$, and the probability that one purchased from B will not fail prior to $x$ hours is a $e^{−0.005}$. Given that a transistor was observed to fail after between $50$ to $60$ hours of operations, what is the probability that it was purchased from $A$?
You want $\mathsf P(A \mid F)$, where $F$ is the event that the resistor failed in the interval, $A,B$ the events that the resistor is from the relevant source. [Note: that the sources are independent does not mean that the events of a resistor coming from a sources are.]
We have been given the CDF for each source and interval $50<x\leqslant 60$, so assuming there was equal (prior)probability of being from each source, this tells us, since $\mathsf P(X\leqslant x;\lambda)$ $=$ $1-e^{-\lambda x}\mathbf 1_{x\in[0;1)}$, that:
$$\mathsf P(A)=\mathsf P(B)=\tfrac 12\\\mathsf P(F\mid A) = -(e^{-0.003\times 60}-e^{-0.003\times 50})\\\mathsf P(F\mid B) = -(e^{-0.005\times 60}-e^{-0.005\times 50})\\\mathsf P(F)=\mathsf P(F\mid A)\mathsf P(A)+\mathsf P(F\mid B)\mathsf P(B)$$
The rest is just an application of Bayes' Rule.