Probability help needed......plzz ?

Question

A fair dice is succesively rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5.
Find the conditional probability density px|y(x|1)

Answer 1

Dear qadir,

The best I can make of what you are asking is that you want to find p(x | 1) where "X | Y" is a subscript of the p. I'll use the notation p(x | 1)_sub(X | Y) to represent that idea. If you can't make your online notation look exactly as originally written, then it would be helpful for you to describe it as clearly as you can in the future (e.g., similar to what I just did). Unfortunately, probability notation can be varied and cumbersome at times.

The way you have characterized the die, there doesn't appear to be a compelling reason for thinking that X and Y are dependent. So, if they are independent, then p(x | 1)_sub(X | Y) would be the same as p(x | n)_sub(X | Y) for any number of rolls n needed to obtain a 5. Further, we could say in this case that p(x | n)_sub(X | Y) is the same a p(x)_sub(X).

Since we're not told the number of faces on the die (the question implies there are at least 6), let's be general and say that the die has k faces, numbered 1 to k, inclusive. This gives us

p(x | 1)_sub(X | Y) = p(x)_sub(X) = (k - 1)^(x - 1) / k^x,

for x a positive integer, otherwise the density is 0.

This density could also be written with fewer exponentiation operations in a form that you might find more intuitive as

(1 / k) ((k - 1) / k)^(x - 1).

In other words, roll x gives you a 6 (success) preceded by x - 1 rolls without a 6 (failures). This is a special case of the negative binomial distribution known as the Pascal distribution.

Answer 2

1/36