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The independent variables X and Y have probability distributions
[tex]P(X=x) = \frac{1}{5} , x = 1, 2, 3, 4, 5\\P(Y=y) = 1/y , y = 2, 3, 6[/tex]

Find [tex]P(Y\ \textgreater \ X)[/tex]

Respuesta :

LammettHash

[tex]Y>X[/tex] for the following [tex](x,y)[/tex]:

(1, 2), (1, 3), (1, 6)

(2, 3), (2, 6)

(3, 6)

(4, 6)

(5, 6)

So we have

[tex]P(Y>X)=P(X=1,Y=2)+P(X=1,Y=3)+\cdots+P(X=5,Y=6)[/tex]

[tex]X,Y[/tex] are independent, so the joint probabilities are

[tex]P(X=x,Y=y)=P(X=x)\cdot P(Y=y)=\dfrac1{5y}[/tex]

Then

[tex]P(Y>X)=\dfrac1{10}+\dfrac2{15}+\dfrac5{30}=\dfrac25[/tex]

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