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I am sorry that I don't get your joke. :tongue: Thank you so much for your help. :laugh:
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Guest replied 
Guest repliedHelp please
I have been working on this question and still have no clues. The solution here is " E(XX>= 3)=(1/p)+2=8, so E[XY=2]=0.2(1)+0.8(8)=6.6".
I don't understand how to get E(XX>= 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.
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Question
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(XY=2).
The answer is 6.6.Tags: None
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