if events X and Y are independent, then for intersection we multiply the probability
P(Y∩X) = P(Y) * P(X)
We know that
[tex]P(Y|X) =\frac{P(YintersectionX)}{P(X)}[/tex]
Now we replace P(Y) * P(X) for P(Y∩X)
[tex]P(Y|X) =\frac{P(Y)*P(X)}{P(X)}[/tex]
Cancel out P(X)
So [tex]P(Y|X) = P(Y)[/tex]
Like that
[tex]P(X|Y) =\frac{P(XintersectionY)}{P(Y)}[/tex]
Now we replace P(X) * P(Y) for P(X∩Y)
[tex]P(X|Y) =\frac{P(X)*P(Y)}{P(Y)}[/tex]
Cancel out P(Y)
So [tex]P(X|Y) = P(X)[/tex]
P(Y | X) = P(Y) and P(X | Y) = P(X) are true
