# Probability involving independent experiments

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 $P(e^{−0.003𝑥})P(e^{−0.005x})$ 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 $e^{−0.003𝑥}$, and the probability that one purchased from B will not fail prior to $x$ hours is a $e^{−0.005𝑥}$. 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 $A$?

• Have you integrated over the time interval? – N. F. Taussig Feb 15 '18 at 21:30
• @N.F.Taussig The CDF is given, so no integration is required. This is a Bayes' Rule problem. – Graham Kemp Feb 15 '18 at 21:31
• PS: The event that the resistor fails and is from A, is not independent from the event that the resistor fails and is from B. Rather they are mutually exclusive (aka disjoint) events. [They cannot both happen.] – Graham Kemp Feb 15 '18 at 21:40

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:
$$\mathsf P(A)=\mathsf P(B)=\tfrac 12\\\mathsf P(F\mid A) = -(e^{-0.003\times 60}-e^{-0.003\times 50})\\\mathsf P(F\mid B) = -(e^{-0.005\times 60}-e^{-0.005\times 50})\\\mathsf P(F)=\mathsf P(F\mid A)\mathsf P(A)+\mathsf P(F\mid B)\mathsf P(B)$$