fy(y) = (fxy(x,y) da 2 = √ ² 1/2 ² 1 (1+3y² ) dx = 1/₁ (1+34²) 5²x dn = 1 / (1+3y² ) [21²] ² = (1+3y²) [4-0] fycy) = 1/2 (1+3y²] 10 ≤ y ≤ a) conditional pdf of x given I fxy (my] fxy (ally) = fyly) x1(1+38²) 거 Z fx11 (21) = 1/2 / 0≤x≤2 b) P(Z
fy(y) = (fxy(x,y) da 2 = √ ² 1/2 ² 1 (1+3y² ) dx = 1/₁ (1+34²) 5²x dn = 1 / (1+3y² ) [21²] ² = (1+3y²) [4-0] fycy) = 1/2 (1+3y²] 10 ≤ y ≤ a) conditional pdf of x given I fxy (my] fxy (ally) = fyly) x1(1+38²) 거 Z fx11 (21) = 1/2 / 0≤x≤2 b) P(Z
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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