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Answer:
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