Problem # 2: An urn contains 3 green balls, 2 red balls and 1 blue ball. Experim

Question

Problem # 2: An urn contains 3 green balls, 2 red balls and 1 blue ball. Experiment # 3: We draw two balls (together) from the urn at random. (a) What is the probability that both balls are red? (b) What is the probability that both balls are green? (c) Let X be the random variable that is 1 if the colors of the two balls drawn in Experiment # 3 are different. Let Y be the random variable that is 1 if the colors of the two balls drawn in Experiment # 2 are the same. Find the covariance of X and Y (d) Interpret your answer to Part (c). (Is the answer what you would expect intuitively? Why?)

Explanation / Answer

Problem No.2

Total number of balls = 6

(a) Pr(Both balls are red) = 2C2/ 6C2 = 1/15

(b) Pr(BOth balls are green) = 3C2/ 6C2  = 3/15 = 1/5

(c) Covaraince (X,Y) = E(XY) -E(X)(Y)

Here f(X) will be 1 when two balls are different. So the options are (1G,1R), (1G,1B), (1R,1B)

P(X) = 1/15 * (3 * 2 + 3 * 1 + 2* 1) = 11/15 for X = 1

= 4/15 for X = 0

P(Y) = 4/15 for Y = 1

= 11/15 for Y = 0

E(X) = 11/15 and E(Y) = 4/15

here E(XY) = 11/15 * 4/15

COva(X,Y) =0

(d) Yes, as both events are mutuall texclusive these events will be indepdent too. So, covariance will be zero here.

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