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# jointexamp - Joint Probability Distributions Examples...

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Joint Probability Distributions – Examples Example 1 A certain market has both an express checkout line and a superexpress checkout line. Let X denote the number of customers in line at the express checkout line at a particular time of day and let Y denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of X and Y is as given in the following table: X / Y 0 1 2 3 0 0.08 0.07 0.04 0.00 1 0.06 0.15 0.05 0.04 2 0.05 0.04 0.10 0.06 3 0.00 0.03 0.04 0.07 4 0.00 0.01 0.05 0.06 : a. What is P ( X = 1 ;Y = 1) , that is, the probability that there is exactly one customer in each line? Solution P ( X = 1 = 1) = p (1 ; 1) = 0 : 15 : b. What is P ( X = Y ) , that is, the probability that the number of customers in the two lines are identical? Solution P ( X = Y ) = p (0 ; 0) + p (1 ; 1) + p (2 ; 2) + p (3 ; 3) = 0 : 08 + 0 : 15 + 0 : 10 + 0 : 07 = 0 : 40 : c. Let A denote the event that there are at least two more customers in one line than in the other line. Express A in terms of X and Y and calculate the probability of this event. Solution A = f ( x;y ) : x ¸ y + 2 g [ f ( ) : y ¸ x + 2 g P ( A ) = p (2 ; 0)+ p (3 ; p (4 ; p (3 ; 1)+ p (4 ; p (4 ; 2)+ p (0 ; p (0 ; 3)+ p (1 ; 3) = 0 : 22 : d. What is the probability that the total number of customers in the two lines is exactly four? At least four? Solution The desired probabilities are: P ( X + Y = 4) = p (1 ; 3) + p (2 ; 2) + p (3 ; 1) + p (4 ; 0) = 0 : 17 P ( X + Y ¸ 4) = P ( X + Y = 4)+ p (4 ; p (4 ; p (4 ; p (3 ; p (3 ; p (2 ; 3) = 0 : 46 : e.

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## jointexamp - Joint Probability Distributions Examples...

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