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Consider the following joint probabilities table for X and Y, where x = 1, 2, 3 and y = 1, 2, 3, 4. X\Y 1 2 3 1 0.05 0.2 0.15 2 3 0.1 0.05 4 0.01 0.02 0.25 0.07 0.01 0.03 0.06 Using the marginal probability functions for X and Y, calculate the following expected values: E[X] = E[Y] = E[4X +5] = E[X + 2Y + 3]: =

Consider the following joint probabilities table for X and Y, where x = 1, 2, 3 and y = 1, 2, 3, 4. X\Y 1 2 3 1 0.05 0.2 0.15 2 3 0.1 0.05 4 0.01 0.02 0.25 0.07 0.01 0.03 0.06 Using the marginal probability functions for X and Y, calculate the following expected values: E[X] = E[Y] = E[4X +5] = E[X + 2Y + 3]: =

A First Course in Probability (10th Edition)
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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Consider the following joint probabilities table for X and Y, where x = 1, 2, 3 and y = 1, 2, 3, 4. X\Y 1 2 3 = 1 0.05 0.2 0.15 Using the marginal probability functions for X and Y, calculate the following expected values: E[X] E[Y] E[4X +5] = E[X + 2Y+3] = = 2 3 4 0.1 0.05 0.01 0.02 0.25 0.07 0.01 0.03 0.06
Transcribed Image Text:Consider the following joint probabilities table for X and Y, where x = 1, 2, 3 and y = 1, 2, 3, 4. X\Y 1 2 3 = 1 0.05 0.2 0.15 Using the marginal probability functions for X and Y, calculate the following expected values: E[X] E[Y] E[4X +5] = E[X + 2Y+3] = = 2 3 4 0.1 0.05 0.01 0.02 0.25 0.07 0.01 0.03 0.06
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Consider the following joint probabilities table for X and Y, where x equals 1 comma space 2 comma space 3 and y equals 1 comma space 2 comma space 3 comma space 4:

X\Y1234
10.050.10.050.01
20.20.020.250.07
30.150.010.030.06
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