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"A fair die is successively rolled. Let X and Y denote,
respectively, the number of rolls necessary to obtain a 6 and a 5.
Find (a) E[X]; (b) E[X|Y=1]; (c) E[X|Y=5]."
X and Y have a geometric distribution, for part b and c we have E[X|Y = 1] = summation P{X=x, Y=1}/P[Y=1} and E[X|Y = 5] = summation P{X=x, Y=5}/P[Y=5}. Can anyone explain how to find P{X=x, Y=1} and P{X=x, Y=5}?

Sagot :

The expected value E[X] = 6 for a geometric distribution

Geometric distribution is a discrete probability distribution. It gives the probability that there are number of  failures before the first success happens.

According to the question,

It is given that

X and Y both follows geometric distribution,

So, Therefore

Probability mass function : p(x) = (1 - p)^(1-x) . p ; x = 1 , 2 , 3 , . . .

where p is probability of getting success

For getting a 6

p = 1/6

and for getting a 5

p = 1/6

So, E[X] = expected value of getting a 6 in a fair die

You should know that the expected value of geometric distribution is

E[X] = 1 / p

Substituting the value of p

E[X] = 1 / (1/6)

=> 6

Similarly you can find rest of parts

To know more about Geometric distribution here

https://brainly.com/question/10164132

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