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hw4 solution

# hw4 solution - 2 a y p(x,y 0 x 1 2 0.30.18.12.6 b...

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2 a. y p(x,y) 0 1 2 3 4 0 .30 .05 .025 .025 .10 .5 x 1 .18 .03 .015 .015 .06 .3 2 .12 .02 .01 .01 .04 .2 .6 .1 .05 .05 .2 b. P(X 1 and Y 1) = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .56 = (.8)(.7) = P(X 1) P(Y 1) c. P( X + Y = 0) = P(X = 0 and Y = 0) = p(0,0) = .30 d. P(X + Y 1) = p(0,0) + p(0,1) + p(1,0) = .53 7 e. p(1,1) = .030 f. P(X 1 and Y 1 = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .120 g. P(X = 1) = p(1,0) + p(1,1) + p(1,2) = .100; P(Y = 1) = p(0,1) + … + p(5,1) = .300 h. P(overflow) = P(X + 3Y > 5) = 1 P(X + 3Y 5) = 1 P*(X,Y)=(0,0) or …or (5,0) or (0,1) or (1,1) or (2,1)± = 1 - .620 = .380 i. The marginal probabilities for X (row sums from the joint probability table) are p x (0) = .05, p x (1) = .10 , p x (2) = .25, p x (3) = .30, p x (4) = .20, p x (5) = .10; those for Y (column sums) are p y (0) = .5, p y (1) = .3, p y (2) = .2. It is now easily verified that for every (x,y), p(x,y) = p x (x) p y (y), so X and Y are independent. 29

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hw4 solution - 2 a y p(x,y 0 x 1 2 0.30.18.12.6 b...

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