The joint probability mass function of X and Y is given byp(1, 1) = 0.15 p(1,2) = 0.1p(2, 1) = 0.1p(1,3) = 0.1p(2, 2) = 0.2p(2, 3) = 0.10.1p(3, 1) = 0.05 p(3,2) = 0.05 p(3,3) = 0.15Compute the following probabilities:P(X+Y > 3)Р(XҮ — 3) —P( > 2) =

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Answered: The joint probability mass function of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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1. Verify whether the following functions can be considered as probability mass functions: (i) P(x = x) = x² + 1 18 (iii) P(X= x) = -, x = 0, 1, 2, 3 (ii) P(X - x) - ²-2, -, x= 1, 2, 3 2x + 1 18 -, x = 0, 1, 2, 3 [Ans.: Yes) [Ans.: No] [Ans.: No]
If the joint pdf of the bivariate random variables X and Y is given by f(x,y) = 6 e-2x - 3Vx >0, y>0. Find the conditional pdf of Y given X Select one: а. f(y|x) =2 e-3y V20 b. f(y|x) = 3 e-2x x20 O C. f(y|x) =2 e-2x ,X20 O d. f(y|x) = 3e-3v V20
b. The table below shows the joint probability mass function for the discrete random variables X and Y. Y = 1 Y = 2 X = 0 X = 1 X = 2 0.07 0.23 0.18 0.05 0.29 0.18 Calculate and interpret the covariance between variables X and Y i.e. Cov(X,Y).

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The joint probability mass function of X and Y is given by p(1,1) = 0.15 p(1,2) = 0.1 p(1,3) = 0.1 p(2,1) = 0.1 p(3, 1) = 0.05 p(3,2) = 0.05 p(3,3) = 0.15 %3D p(2,2) = 0.2 p(2,3) = 0.1 %3D %3D %3D Compute the following probabilities: P(X+Y > 3) = P(XY = 3) = %3D P(\ > 2) =