Let X be the number of 1's and Y the number of 2's that occur in n rolls of a fair die. Compute Cov(X,Y).
Do I use ?
A fair die is rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find E[X|Y=5].
Let X be the number of 1's and Y the number of 2's that occur in n rolls of a fair die. Compute Cov(X,Y).
Do I use ?
A fair die is rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find E[X|Y=5].
Hello,
Why not ?
You can compute E[X] and E[Y] I guess ^^ (same binomial distributions)
For the distribution , note that it's equivalent to saying "what is the probability to get l 2's, while there already had k 1's ?
You can also see that l can get values only between 0 and n-k.
And that it follows a binomial distribution (n-k,1/5) ~ Z
So we actually have :
But looks like a lot , doesn't it ?
Can you try to do this ?
Yes, but the sum isn't exactly this !
There is the sum , which depends on k, so you can't just calculate the two sums separately.
I'm sorry I should've added brackets :
Actually, do you understand why , where Z follows a binomial distribution (n-k,1/5) ?However I'm stuck with , especially the (n-k) part
Do it with words : P(Y=l|X=k) is the probability that there are l 2's, while there are k 1's, among n numbers. So it means that there already have k 1's, leaving us with n-k 'undefined' numbers.
But, there can't be any 1's left in these n-k, since they're included in the k numbers (X=k is the total of 1's among the n numbers). So there are 5 remaining possibilities : {2,3,4,5,6}
Thus, P(Y=l|X=k) is equal to P(Z=l), where Z follows a binomial distribution (n-k,1/5)
Does it look better this way ?
Now the sum :
Then you're left with :
But
And
Finally,
And you get the desired result