Let Z = XY. 1. Find the expectation E[Z]. E[Z] =18.6667
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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How was the correct answer of 18.6667 attained?
![Suppose the joint PMF of the random variables X and Y is
P(X = x, Y = y)
Let Z = XY.
1. Find the expectation E[Z].
E[Z] =
= 18.6667
=
x+y
81
if(x, y) {5,6,7} × {2,3,4},
0 Otherwise.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1467f162-7185-45ab-925a-2589d3c8cce7%2F6b5a6dd5-0d01-4d5e-8ad0-2bd332695e99%2Frwq9xio_processed.png&w=3840&q=75)
Transcribed Image Text:Suppose the joint PMF of the random variables X and Y is
P(X = x, Y = y)
Let Z = XY.
1. Find the expectation E[Z].
E[Z] =
= 18.6667
=
x+y
81
if(x, y) {5,6,7} × {2,3,4},
0 Otherwise.
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