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ma316f03hw3

# ma316f03hw3 - MA 316 MATHEMATICAL PROBABILITY ASSIGNMENT#3...

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MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #3: SOLUTIONS: SECTIONS 1.5, 1.6 FALL 2003 1.48: For part a), if f ( x ) = c ± 2 3 x for x = 1 , 2 , 3 ,... then we need c x =1 ( 2 3 ) x = 1 . Since in general x =0 r x = 1 / (1 - r ) then x =1 r x = 1 / (1 - r ) - 1 so that c x =1 ( 2 3 ) x = c (1 / (1 - 2 / 3) - 1) - 1 = c · 2 = 1 which implies that c = 1 / 2. For part b) we need that c + 2 c + 3 c + 4 c + 5 c + 6 c = 1 so c = 1 / 21. 1.49: If f ( x ) = x/ 15 for x = 1 , 2 , 3 , 4 , 5 , 6 and is zero elsewhere then P ( X = 1 orX = 2) = P (1 / 2 < X < 5 / 2) = P (1 X 2) = 1 / 15 + 2 / 15 = 1 / 5 . 1.51: For part a), if X is a random variable that denotes the number of hearts in a 5 card hand then P ( X = k ) = ( 13 k )( 39 5 - k ) ( 52 5 ) . For part b, the probability that X = 1 or X = 0 is then P ( X = 0 or X = 1) = ( 13 0 )( 39 5 ) + ( 13 1 )( 39 4 ) ( 52 5 ) . 1.53:
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