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# asst456 - York University MATH 2030 3.0(Elementary...

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York University MATH 2030 3.0 (Elementary Probability) Fall 2011 Assignment 4 – Solutions, Oct 2011 § 3.1 No. 9 X = 2 if the 2nd draw matches the first, which has probability 1 7 . X = 3 if the 2nd draw differs from the first, and the third matches the 1st or 2nd. This has probability 6 7 × 2 6 = 2 7 . Likewise P ( X = 4) = 6 7 × 4 6 × 3 5 = 12 35 . If the first 4 draws differ, then the 5th must match one of them, so P ( X = 5) = 1 - 1 7 - 2 7 - 12 35 = 8 35 . We obtain x 2 3 4 5 P ( X = x ) 1 7 2 7 12 35 8 35 § 4.1 No. 3abcd (a) We want 1 = -∞ f ( x ) dx = 1 0 cx (1 - x ) dx = c 1 0 ( x - x 2 ) dx = c x 2 2 - x 3 3 1 0 = c ( 1 2 - 1 3 ) = c 6 So we take c = 6. (b) Using part (a), P ( X 1 2 ) = 1 / 2 -∞ f ( x ) dx = 1 / 2 0 6 x (1 - x ) dx = 1 / 2 0 (6 x - 6 x 2 ) dx = 3 x 2 - 2 x 3 1 / 2 0 = 3 4 - 1 4 = 1 2 (c) P ( X 1 3 ) = 1 / 3 -∞ f ( x ) dx = 1 / 3 0 6 x (1 - x ) dx = 1 / 3 0 (6 x - 6 x 2 ) dx = 3 x 2 - 2 x 3 1 / 3 0 = 1 3 - 2 27 = 7 27 = 0 . 2593 (d) P ( 1 3 < X 1 2 ) = P ( X 1 2 ) - P ( X 1 3 ) = 1 2 - 7 27 = 13 54 = 0 . 2407 § 4.5 No. 2a Let X Bin(3 , 1 2 ). Using the formula P ( X = k ) = ( 3 k ) (1 / 2) k (1 / 2) 3 - k we compute that P ( X = 0) = 1 / 8, P ( X = 1) = 3 / 8, P ( X = 2) = 3 / 8, and P ( X = 3) = 1 / 8. So the cdf F ( x ) of X is a step function. F ( x ) = 0 , x < 0 , ; F ( x ) = 1 / 8 , x [0 , 1); F ( x ) = 1 / 8 + 3 / 8 = 4 / 8 , x [1 , 2); F ( x ) = 4 / 8 + 3 / 8 = 7 / 8 , x [2 , 3); F ( x ) = 7 / 8 + 1 / 8 = 1 , x 3. You draw a pic according to this. Note that the 1

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2 jump height at each jump point x = 0 , 1 , 2 , 3 follows from the defining pr at each x = 0 , 1 , 2 , 3. Note that at the jump points x = 0 , 1 , 2 , 3 the cdf takes the upper value, so that the function is right-continuous with left-limits (this is the defining property of a cdf). § 4.5 No. 5 F ( x ) = P ( X x ) = x -∞ f ( t ) dt = x -∞ e t 2 dt, x < 0 0 -∞ e t 2 dt + x 0 e - t 2 dt, x 0 = x -∞ e t 2 , x < 0 0 -∞ e t 2 + x 0 - e - t 2 , x 0 = e x 2 , x < 0 1 - e - x 2 , x 0 § 4.5 No. 6ab (a) Because P ( X = 1 / 2) = 0 we have that P ( X 1 / 2) = 1 - P ( X < 1 / 2) = 1 - F (1 / 2) = 1 - 1 8 = 7 / 8.
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asst456 - York University MATH 2030 3.0(Elementary...

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