.oo du LTE 4:12 PM courses1.webwork.maa.org webwork /petroleuminst-math045/ 3396

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.oo du LTE 4:12 PM courses1.webwork.maa.org webwork /petroleuminst-math045/ 3396. ch5/3 Ch5: Problem3 Previous Problem Problem List Next Problem (1 point) The joint probability mass function of X and Y is given by p(1,1)= 0.35 p(1,2)= 0.05 p(1,3)= 0.1 p(2,1)= 0.05 p(2,2)= 0.2 p(2,3)=0.1 p(3,1) = 0.05 p(3,2)= 0.05 p(3,3)= 0.05 (a) Compute the conditional mass function of Y given X = 1: P(Y = 1|X = 1) = P(Y = 2|X = 1) = P(Y = 31X = 1) = (b) Are X and Y independent? (enter YES or NO) Note: You can earn partial credit on this problem. Preview My Answers Submit Answers

Explanation / Answer

P(X = 1) = P(1, 1) + P(1, 2) + P(1, 3) = 0.35 + 0.05 + 0.1 = 0.5

A) P(Y = 1 | X = 1) = P(1,1) / P(X =1) = 0.35 / 0.5 = 0.7

P(Y = 2 | X = 1) = P(1,2) / P(X = 1) = 0.05/0.5 = 0.1

P(Y = 3 | X = 1) = P(1,3) / P(X = 1) = 0.1 / 0.5 = 0.2

B) P(Y = 1) = P(1,1) + P(2,1) + P(3,1) = 0.35 + 0.05 + 0.05 = 0.45

Since, P(Y = 1 | X = 1) is not equal to P(Y = 1), X and Y are not independent.

The answer is NO

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