Urgent s2 help

Jean regularly takes a break from work to go to the post office. The amount of time Jean waits in the queue to be served at the post office has a continuous uniform distribution between 0 and 10 minutes.

Jean is in the queue when she receives a message that she must return to work for an urgent meeting. She can only wait in the queue for a further 3 minutes. Given that Jean has already been queuing for 5 minutes,
(d) find the probability that she must leave the post office queue without being served.

can someone pls help me with this part??
why can't I just do p(5<X<8)

Because that would be the probability she gets served between 5 and 8 minutes after arriving, and not does it take into account that she's waited 5 minutes already.

Two things:

She's been there 5 min and can wait a further 3. So, she won't get served if X>8.

So, we're looking initially at P(X>8).

But also we're told she's waited 5 minutes already - that a given - , so we're actually looking for P(X > 8 | X > 5).

okay I got that but why did we use p(X>5) not p(X<5)? I mean she already waited 5 minutes right? @ghostwalker

The r.v. X is the amount of time since she arrived. We know she's been there for 5 minutes, hence X is greater than 5.

ah okay that makes sense! can I just put greater than every time I see already?

That question is far too general to give you an answer.

You need to look at what the r.v. represents. In this case, X is the amount of time she waits in a queue to be served. You know she's been there 5 minutes already, so X>5.

The only confusion, I can see that might arise is in thinking X represents the time left, which it doesn't. And you can deal with that confusion by going back to look at what X represents.