# Uniform distribution

1. Dec 28, 2011

### jsmith613

1. The problem statement, all variables and given/known data
http://www.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/S2/S2%202008-06.pdf [Broken]

Question 1(d)
2. Relevant equations

3. The attempt at a solution
So I know this is a conditional probability question.
Now I would have said
P(X>8) / P(X=5)
because it is probability I will NOT be served (i.e: time is greater than 8 mins) given I have already waited 5 mins
this gives me infinity because x/0 = inifinty

How do I approach this question?

Last edited by a moderator: May 5, 2017
2. Dec 28, 2011

### Ray Vickson

No: because she has already waited 5 minutes, we know that X >= 5. You need P{X > 8}/P{X >= 5}.

RGV

Last edited by a moderator: May 5, 2017
3. Dec 29, 2011

### jsmith613

see thats the prob I can't see why its x≥5 because she has waited 5 mins (x=5)
X = the amount of time Jean waits in the queue?

Last edited by a moderator: Dec 29, 2011
4. Dec 29, 2011

### Ray Vickson

X = amount of time that Jean would spend in the queue if she did not have to leave before being served; this is the random quantity representing the times she did spend in the queue on every day in the past. So, X is uniform from 0 to 10. On this one occasion we observe that after 5 minutes she is still in the queue, so on this one occasion, X >= 5. (X is not exactly equal to 5, because if it was she would not need to wait any longer; and, of course, X is not less than 5.) On this one occasion, the amount of time she spends in the queue is X (if X <= 8) and is 8 (if X > 8).

Last edited by a moderator: Dec 29, 2011
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