prob7 - Mathematical Systems Probability Solutions by...

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Mathematical Systems Probability Solutions by Bracket A First Course in Probability Chapter 6—Problems 1. Two fair dice are rolled. Find the join probability mass function of X and Y when a. X is the largest value obtained on any die and Y is the sum of their values. We have p ( x, y ) as given by the table below: x y 2 3 4 5 6 7 8 9 10 11 12 1 1 / 36 - - - - - - - - - - 2 - 2 / 36 1 / 36 - - - - - - - - 3 - - 2 / 36 2 / 36 1 / 36 - - - - - - 4 - - - 2 / 36 2 / 36 2 / 36 1 / 36 - - - - 5 - - - - 2 / 36 2 / 36 2 / 36 2 / 36 1 / 36 - - 6 - - - - - 2 / 36 2 / 36 2 / 36 2 / 36 2 / 36 1 / 36 b. X is the value on the first die and Y is the larger of the two values. We have p ( x, y ) as given by the table below: x y 1 2 3 4 5 6 1 1 / 36 1 / 36 1 / 36 1 / 36 1 / 36 1 / 36 2 - 2 / 36 1 / 36 1 / 36 1 / 36 1 / 36 3 - - 3 / 36 1 / 36 1 / 36 1 / 36 4 - - - 4 / 36 1 / 36 1 / 36 5 - - - - 5 / 36 1 / 36 6 - - - - - 6 / 36 c. X is the smallest and Y is the largest value obtained on the dice. Note that the event corresponding to p ( x, y ) occurs precisely when we roll ( x, y ) or ( x, y ) if x < y , in which case p ( x, y ) = 2 36 , or p ( x, y ) = 1 36 if x = y . 7. Consider a sequence of independent Bernoulli trials, each of which is a success with probability p . Let X 1 be the number of failures preceding the first success, and let X 2 be the number of failures between the first two successes. Find the joint mass function of X 1 and X 2 . Note that X 1 and X 2 are independent, and thus P ± X 1 = x 1 , X 2 = x 2 ² = P ± X 1 = x 1 ² P ± X 2 = x 2 ² . Let X be a geometrically distributed random variable with parameter p . Then P ± X 1 = x 1 ² = P ± X = x 1 + 1 ² = (1 - p ) x 1 p , and similarly for X 2 . Thus p ( x 1 , x 2 ) = (1 - p ) x 1 + x 2 p 2 . 1
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Mathematical Systems Probability 8. The joint probability density function of X and Y is given by f ( x, y ) = c ( y 2 - x 2 ) e - y - y x y, 0 < y < a. Find c . We integrate over the entire region to obtain 1 = Z Z C f ( x, y ) dx dy = c Z 0 Z y - y ( y 2 - x 2 ) e - y dx dy = c Z 0 e - y Z y - y ( y 2 - x 2 ) dx dy = c Z 0 (2 y 3 - 2 3 y 3 ) dy = c Z 0 e - y (2 y 3 - 2 3 y 3 ) dy = 4 3 c Z 0 y 3 e - y dy = 4 3 c · Γ(3 + 1) = 4 3 c · 3! = 8 c Thus c = 1 8 . b. Find the marginal densities of X and Y . We have f X ( x ) = Z -∞ f ( x, y ) dy = Z | x | c ( y 2 - x 2 ) e - y dy = c ( x 2 - y 2 - 2 y - 2) e - y | y = | x | = 1 8 (2 | x | + 2) e -| x | = 1 4 ( | x | + 1) e -| x | . Similarly f Y ( y ) = Z -∞ f ( x, y ) dx = Z y - y c ( y 2 - x 2 ) e - y dx = ce - y Z y - y ( y 2 - x 2 ) dx = ce - y (2 y 3 - 2 3 y 4 ) = 1 8 · e - y · 4 3 y 3 = 1 6 y 3 e - y . c. Find E [ X ]. E [ X ] = Z -∞ xf X ( x ) dx = 1 4 Z -∞ x ( | x | + 1) e -| x | dx = 1 4 Z 0 -∞ x ( - x + 1) e x dx + 1 4 Z 0 x ( x + 1) e - x dx = 1 4 ± e x ( - x 2 + 3 x - 3) ² 0 x = -∞ + 1 4 ± - e - x ( x 2 + 3 x + 3) ² x =0 = - 3 4 + 3 4 = 0 2
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Mathematical Systems Probability 16. Suppose that n points are independently chosen at random on the perimeter of a circle, and we want the probability that they all lie in some semicircle. (That is, we want the probability that there is
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prob7 - Mathematical Systems Probability Solutions by...

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