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Determine E (Y) and V (Y) y f(x, y) -1.0 -2 1/3 -0.5 -1 1/6 0.5 1 1/6 1.0 1/3 O a. E(Y) = 0 V(V) = 2.50 O b. None O c. E(Y) = 1/4 V(V) = 1.6875 %3D O d. E(Y) = 0 V(V) = 3.00 %3D

Determine E (Y) and V (Y) y f(x, y) -1.0 -2 1/3 -0.5 -1 1/6 0.5 1 1/6 1.0 1/3 O a. E(Y) = 0 V(V) = 2.50 O b. None O c. E(Y) = 1/4 V(V) = 1.6875 %3D O d. E(Y) = 0 V(V) = 3.00 %3D

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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Determine E (Y) and V (Y) y f(x, y) -1.0 -2 1/3 -0.5 -1 1/6 0.5 1 1/6 1.0 1/3 O a. E(Y) = 0 V(V) = 2.50 O b. None O c. E(Y) = 1/4 V(V) = 1.6875 %3D O d. E(Y) = 0 V(V) = 3.00
Transcribed Image Text:Determine E (Y) and V (Y) y f(x, y) -1.0 -2 1/3 -0.5 -1 1/6 0.5 1 1/6 1.0 1/3 O a. E(Y) = 0 V(V) = 2.50 O b. None O c. E(Y) = 1/4 V(V) = 1.6875 %3D O d. E(Y) = 0 V(V) = 3.00
Conditional probability distribution X given that Y = 1 f(x, y) -1.0 -2 1/4 -0.5 -1 1/2 0.5 1 1/8 1.0 2 1/8 fxy (x, 1) fxjr (x) = fr(1) a. None b. 1/4 O c. 1 d. 1/8
Transcribed Image Text:Conditional probability distribution X given that Y = 1 f(x, y) -1.0 -2 1/4 -0.5 -1 1/2 0.5 1 1/8 1.0 2 1/8 fxy (x, 1) fxjr (x) = fr(1) a. None b. 1/4 O c. 1 d. 1/8
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