131 STAT 155 Chapter 5 Joint Probability Distributions and Random Samples

# 131 stat 155 chapter 5 joint probability

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131 STAT 155 Chapter 5: Joint Probability Distributions and Random Samples Section 5.1: Jointly Distributed Random Variables__________________________________ Let X and Y be two discrete rv’s defined on the sample space S of an experiment. The joint probability mass function p(x, y) is defined for each pair of numbers (x, y) by p(x, y) = P(X = x and Y = y). It must be the case that p(x, y) 0 and x y p(x,y) = 1.   Now let A be any set consisting of pairs of (x, y) values (e.g., A = {(x, y): x + y = 5} or {(x, y): max(x, y) 3}). Then the probability P[(X,Y) A] is obtained by summing the joint pmf over pairs in A: P[(X, Y) A] = (x ,y) A p (x , y)   The marginal probability mass functions of X and of Y, denoted by p x (x) and p y (y), respectively, are given by p x (x) = y p(x, y) p y (y) = p(x, y) x 132 5.2 When an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight is checked to see whether it is properly aimed. Let X denote the number of headlights that need adjustment, and let Y denote the number of defective tires. a.If X and Y are independent with px(0) = .5, px(1) = .3, px(2) = .2, and py(0) = .6, py(1) = .1, py(2) = py(3) = .05, py(4) = .2, display the joint pmf of (X, Y) in a joint probability table. b.Compute P(X 1 and Y 1) from the joint probability table, and verify that it equals the product P(X 1)P(Y 1). c.What is P(X + Y = 0), the probability of no violations? d.Compute P(X + Y 1). 133 d) = 135  • • • 