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ISE361_HW5answers_F03

ISE361_HW5answers_F03 - Homework Five 2 a p(x,y x 0| 1| 2|...

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Homework Five Fall 03 2) a. y p(x,y) 0 1 2 3 4 x 0 | .30 .05 .025 .025 .10 .5 1 | .18 .03 .015 .015 .06 .3 2 | .12 .02 .01 .01 .04 .2 .60 .10 .05 .05 .2 1.0 b. P(X<=1 and Y<=1) = p(0,0) + p(0,1) + p(1,0) + p(1,1) = 0.56 P(X<=1) x P(Y<=1) = (.8)(.7) = 0.56 6) a. p(4,2) = P(Y=2|X=4) P(X=4) = [ C 2,4 (.6) 2 (.4) 2 ] (.15) = 0.0518 b. P(X=Y) = p(0,0) + p(1,1) + p(2,2) + p(3,3) + p(4,4) = .1 + (.2)(.6) + (.3)(.6) 2 + (.25)(.6) 3 + (.15)(.6) 4 = 0.4014 8) a. numerator = C 3,8 C 2,10 C 1,12 = (56)(45)(12) = 30,240 denominator = C 6,30 = 593,775; p(3,2) = (30,240/593,775) = 0.0509 9) a. 30 30 k(x 2 +y 2 )dxdy = 1 k = 3 / 380,000 20 20 26 26 b. P(X<26 and Y<26) = k (x 2 +y 2 )dxdy = (38,304)(k) = 0.3024 20 20 10) a. f (x,y) = 1 5 <= x <= 6; 5 <= y <= 6 0 otherwise b. P(5.25<=X<=5.75, 5.25<=Y<= 5.75) = P(5.25<=X<=5.75) x P(5.25<=Y<=5.75) =(.5)(.5) = 0.25 (by independence) 12) a. P(X>3) = 3 4 0 4 x e -x(1+y) dydx = 3 e -x dx = 0.05 b. The marginal pdf of X is 0 xe -x(1+y) dy = e -x = for 0 <= x; that of Y is 3 xe -x(1+y) dx = 1 / (1+y) 2 for 0<=y. It is now clear that f(x,y) is not the product of the marginal pdf’s; so the 2 RV’s are not independent c. P(at least one exceeds 3) = 1 – P(X<= 3 & Y<=3) = 1 - 0 3 0 3 xe -x(1+y) dydx = 1 - 0 3 0 3 xe -x e -xy dy = 1 - 0 3 e -x (1-e -3x )dx = e -3 + .25 - .25 e -12 = .300 - 1 -
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Homework Five 26) Revenue = 3X + 10Y, so E (revenue) = E(3X + 10Y) = {x = 0to5} {y= 0to2} [3x + 10y] p(x,y) = 0 p(0,0) + …+ 35 p(5,2) = 15.4
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