Question

Thank you

First, I guess you didnt get my joke.

Second, this problem has been discussed in great detail here:

http://www.actuary.com/actuarial-dis...ad.php?t=14540

I am sorry that I don't get your joke. :tongue: Thank you so much for your help. :laugh:

Help please

I have been working on this question and still have no clues. The solution here is " E(X|X>= 3)=(1/p)+2=8, so E[X|Y=2]=0.2(1)+0.8(8)=6.6".
I don't understand how to get E(X|X>= 3)=(1/p)+2=8??? I though the p for X should be 1/5 since Y is the number of rolls to get a 6 and Y=2. Also, where does "2" come from???
For 0.2 and 0.8, is it because the probability of the first time to get a 5 is 1/5=0.2 and the probability after the 3rd time is [1-(1/5)]=0.8 under the condition of Y=2? I just want to make sure my thoughts are right.

A fair die is rolled repeatedly. Let X be the number of rolls needed to obtain a 5 and Y be the number of rolls needed to obtain a 6.
Calculate E(X|Y=2).

The answer is 6.6.