# 180asolns6 - 180A HW 6 Solutions 1 Consider an urn with 1...

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180A HW 6 Solutions November 19, 2006 1) Consider an urn with 1 red, 2 green and 3 blue balls. 5 balls are drawn from that urn without replacement. Let X be the number of red balls picked and Y the number of green balls picked. Find the joint distribution of X and Y . Answer We note that 0 X 1 and 0 Y 2 , and that X + Y 2 . Thus for any x, y such that 0 x 1 and 2 - x y 2 we have that f X,Y ( x, y ) = P ( X = x, Y = y ) = ( 1 x )( 2 y )( 3 5 - y - x ) ( 6 5 ) and for any other pair ( x, y ) with x + y < 2 or x > 1 or y > 2 that f X,Y ( x, y ) = 0 . 2) Consider an urn with 1 red, 2 green and 3 blue balls. 100 balls are drawn from that urn with replacement. What is the probability that: (a) There are 15 red, 35 green and 50 blue balls? (b) All balls are of the same color? (c) There are more blue balls than red and green balls together? Answer For ( a ) note that there are 100! 15!35!50! possible drawing sequences, and the probability of any given one is ( 1 6 ) 15 ( 1 3 ) 35 ( 1 2 ) 50 so if X , Y and Z are the number of red, blue and green balls respectively then P ( X = 15 , Y = 35 , Z = 50) = 100! 15!35!50! 1 6 15 1 3 35 1 2 50 For ( b ) note that P ( all balls are the same color ) = P ( X = 100) + P ( Y = 100) + P ( Z = 100) = 1 6 100 + 1 3 100 + 1 2 100 For ( c ) we note that B = 100 - ( R + G ) and there are more blue balls drawn than red and green together if B > 50 . Letting Z , as before, denote the number of blue balls, we note that Z Bin (100 , 0 . 5) and hence P ( Z > 50) = 100 X k =51 100 k 2 - 100 1

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We can indeed compute this exactly; noting that 100 X
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