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It is known that X = 1,2,3 and 4 = 1,2,3 are two variables with the probability Mass function p(x=x₁y+y) expressed as p(x,y) which is given as follows: P (1.4) 1 P (2, 1) = 1 P (3.1)= 1 3 P (1,2) Based 1 g P(1,3) O P (212) = 0 р 6 a) Find P(X=X |Y= 1)! b) Find P (Y=y | X= 1)! P (312)= 1 18 P (313) 1 on the probability information above 9 c show that variable X and variable + are not Independent !

It is known that X = 1,2,3 and 4 = 1,2,3 are two variables with the probability Mass function p(x=x₁y+y) expressed as p(x,y) which is given as follows: P (1.4) 1 P (2, 1) = 1 P (3.1)= 1 3 P (1,2) Based 1 g P(1,3) O P (212) = 0 р 6 a) Find P(X=X |Y= 1)! b) Find P (Y=y | X= 1)! P (312)= 1 18 P (313) 1 on the probability information above 9 c show that variable X and variable + are not Independent !

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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It is known that X = 1,2,3 and 4 =1,2,3 are two variables with the probability Mass function P(x=x₁yty expressed as as p(x,y) which is given as follows: p (2₁1) = 1/1/20 스 35 P(3.1) 1 3 P (212) = 0 P (312) = 1 P (313) P (1,2) g P(1.3) = 0 D(2-3) = 2 6 Based on the probability information above a) Find P (X=X |Y= 1)! b) Find P (Y=y | X= 1)! 1 9 show that variable x and variable & are not Independent !
Transcribed Image Text:It is known that X = 1,2,3 and 4 =1,2,3 are two variables with the probability Mass function P(x=x₁yty expressed as as p(x,y) which is given as follows: p (2₁1) = 1/1/20 스 35 P(3.1) 1 3 P (212) = 0 P (312) = 1 P (313) P (1,2) g P(1.3) = 0 D(2-3) = 2 6 Based on the probability information above a) Find P (X=X |Y= 1)! b) Find P (Y=y | X= 1)! 1 9 show that variable x and variable & are not Independent !
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