The joint probability mass function of XX and YY is given by p(1,1)=0p(2,1)=0.1p(3,1)=0.05p(1,2) =0.05p(2,2)=0.3p(3,2)=0.1p(1,3)=0.0 5p(2,3)=0.1p(3,3)=0.25 p(1,1)=0p(1,2)=0.05p(1,3)=0.05p(2,1 )=0.1p(2,2)=0.3p(2,3)=0.1p(3,1)=0.0 5p(3,2)=0.1p(3,3)=0.25 (a) Compute the conditional mass function of YY given X=3X=3: P(Y=1|X=3)

P(Y=1|X=3)=0.125

Step-by-step explanation:

Given :

p(1,1)=0  

p(2,1)=0.1

p(3,1)=0.05

p(1,2)=0.05

p(2,2)=0.3

p(3,2)=0.1

p(1,3)=0.05

p(2,3)=0.1

p(3,3)=0.25

Now we are supposed to find the conditional mass function of Y given X=3 :  P(Y=1|X=3)

P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)

P(X=3)=p(3,1) +p(3,2) +p(3,3)

P(X=3)=0.05+0.1+0.25=0.4

Hence P(Y=1|X=3)=0.125

is this part of a question or something

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6