The joint probability mass function of X and Y is given by p(1, 1) — 0.55 рp(1,2) — 0.05 p(1,3) — 0.05 Р(2, 1) — 0.1 p(2, 2) — 0 Р(3, 1) — 0.1 p(2, 3) = 0.05 p(3, 2) = 0.05 p(3, 3) = 0.05 %3D %3D (a) Compute the conditional mass function of Y given X = 1: P(Y = 1|X = 1) = P(Y = 2|X = 1) =| %3D %3D %3D %3D P(Y = 3|X = 1) = %3D (b) Are X and Y independent? (enter YES or NO) (c) Compute the following probabilities: P(X+Y > 2) : P(XY = 2) = P( > 1) =|

The joint probability mass function of X and Y is given by p(1, 1) — 0.55 рp(1,2) — 0.05 p(1,3) — 0.05 Р(2, 1) — 0.1 p(2, 2) — 0 Р(3, 1) — 0.1 p(2, 3) = 0.05 p(3, 2) = 0.05 p(3, 3) = 0.05 %3D %3D (a) Compute the conditional mass function of Y given X = 1: P(Y = 1|X = 1) = P(Y = 2|X = 1) =| %3D %3D %3D %3D P(Y = 3|X = 1) = %3D (b) Are X and Y independent? (enter YES or NO) (c) Compute the following probabilities: P(X+Y > 2) : P(XY = 2) = P( > 1) =|

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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The joint probability mass function of X and Y is given by
Р(1, 1) — 0.55 p(1,2) — 0.05 р(1,3) — 0.05
p(2, 3) = 0.05
=0.05 p(3, 3) = 0.05
p(2, 1) = 0.1
Р(3, 1) — 0.1 p(3, 2)
Р(2, 2) — 0
%3D
(a) Compute the conditional mass function of Y given X =1: P(Y = 1|X = 1) =
Р(Y — 2|X — 1)
%3D
%3D
%3D
P(Y = 3|X = 1) =
(b) Are X and Y independent? (enter YES or NO)
(C) Compute the following probabilities:
P(X+Y > 2) =
P(XY = 2)
P( > 1) =
%3D

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