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Suppose X and Y are independent Binomial random variables, each with n=3 and p=9/10. a. Find...

Suppose X and Y are independent Binomial random variables, each with n=3 and p=9/10.

a. Find the probability that X and Y are equal, i.e., find P(X=Y).

b. Find the probability that X is strictly larger than Y, i.e., find P(X>Y).

c. Find the probability that Y is strictly larger than X, i.e., find P(Y>X).

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Answer #1

a) The probability here is computed as:

P(X = Y) = P(X = Y = 0) + P(X = Y = 1) + P(X = Y = 2) + P(X = Y = 3)

P(X = Y) = 0.1^3*0.1^3 + (3*0.9*0.1^2)*(3*0.9*0.1^2) + (3*0.9^2*0.1)*(3*0.9^2*0.1) + 0.9^3*0.9^3

P(X = Y) = 0.5912

Therefore 0.5912 is the required probability here.

b) Now there are two possibilities here, either X > Y or Y > X in case where X is not equal to Y.

P(X not equal to Y) = 1 - P(X = Y) = 1 - 0.5912 = 0.4088

Therefore due to symmetry we would have: P(X > Y) = P(Y > X) = 0.4088 / 2 = 0.2044

Therefore 0.2044 is the required probability here.

c) As explained in the previous part itself, P(X > Y) = P(Y > X) = 0.2044

Therefore 0.2044 is the required probability here.

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