Homework Help is Here – Start Your Trial Now!
Math
Probability
The joint probability mass function of X and Y is given by p(1, 1) = 0.05 p(1,2) = 0.1 p(1, 3) = 0.05 p(2, 1) = 0.1 p(2, 2) = 0.25 p(2, 3) = 0.1 p(3, 1) = 0.05 p(3,2) = 0.1 p(3,3) = 0.2 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) = .2222 %3D P(Y = 2|X = 2) = .5556 P(Y = 3|X = 2) = .2222 (b) Are X and Y independent? (enter YES or NO) NO (c) Compute the following probabilities: P(X+Y > 3) = .65 P(XY = 3) = .1 P( > 2) = | .05

The joint probability mass function of X and Y is given by p(1, 1) = 0.05 p(1,2) = 0.1 p(1, 3) = 0.05 p(2, 1) = 0.1 p(2, 2) = 0.25 p(2, 3) = 0.1 p(3, 1) = 0.05 p(3,2) = 0.1 p(3,3) = 0.2 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) = .2222 %3D P(Y = 2|X = 2) = .5556 P(Y = 3|X = 2) = .2222 (b) Are X and Y independent? (enter YES or NO) NO (c) Compute the following probabilities: P(X+Y > 3) = .65 P(XY = 3) = .1 P( > 2) = | .05

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...
See similar textbooks
icon
Related questions
Question
Please answer the incorrect question marked in red. Thank you!
The joint probability mass function of X andY is given by p(1, 1) = 0.05 p(1,2) = 0.1 p(1, 3) = 0.05 p(2, 1) = 0.1 p(2, 2) = 0.25 p(2,3) = 0.1 p(3, 1) = 0.05 p(3,2) = 0.1 p(3,3) = 0.2 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) = .2222 P(Y = 2|X = 2) = .5556 P(Y = 3|X = 2) = .2222 (b) Are X andY independent? (enter YES or NO) NO (c) Compute the following probabilities: P(X+Y > 3) = .65 P(XY = 3) = .1 P( > 2) = .05
Transcribed Image Text:The joint probability mass function of X andY is given by p(1, 1) = 0.05 p(1,2) = 0.1 p(1, 3) = 0.05 p(2, 1) = 0.1 p(2, 2) = 0.25 p(2,3) = 0.1 p(3, 1) = 0.05 p(3,2) = 0.1 p(3,3) = 0.2 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) = .2222 P(Y = 2|X = 2) = .5556 P(Y = 3|X = 2) = .2222 (b) Are X andY independent? (enter YES or NO) NO (c) Compute the following probabilities: P(X+Y > 3) = .65 P(XY = 3) = .1 P( > 2) = .05
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer