# condition probability quick question

Suppose I had a batch of old computer screens. 30% of these are bad 70% are good. The lifetime of both of these screens are indepdent exponential random variables. The mean for the bad screens is 150 days, and the mean for the good screens are 300 days.

Let X be a r.v. = the lifetime of the bad screens

Let Y be a r.v. = the lifetime of the good screens

let Z be a r.v. = lifetime of a randomly selected screen.

I have found the pdf of Z

I am asked: given that i choose a screen at random, and the chosen one is still working after 400 days, what is the probability it came from the "bad" batch?

I am asking here on how to interpret this question. I figured the "still working after 400 days" means "Z>=400", but what is being asked?

I think it's P(something | Z>= 400) but what is something?

thanks

• use tex please. – Seyhmus Güngören Feb 12 '15 at 1:31
• This informal phrasing of the problem forces you to think carefully about the setup. Literally the problem asks "the probability it came from the 'bad' batch", and from what you've written just before that, the screen was chosen "at random". You need a new random variable to represent that outcome (which is binary). – hardmath Feb 12 '15 at 1:44

$X$ a screen is good, $Y=X'$ a screen is bad, $Z$ screen is still operating after 300 days.

What you need: $P(Y|Z)$

How to get it: Bayes' Theorem