# prob7 - Mathematical Systems Probability Solutions by...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

prob7 - Mathematical Systems Probability Solutions by...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online