Let X have the pmf p(x) = (1/3)(2/3)^x, x = 0, 1, 2, 3, ..., zero elsewhere. Find conditional pmf of X given that X ? 2.
Added by Luis P.
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We can do this by using Bayes' theorem: P(X>=2 | X=x) = P(X=x | X>=2) * P(X>=2) / P(X=x) We know that P(X>=2) = P(X=2) + P(X=3) = (1/3)*(2/3) + (1/3)*(2/3)^2 = 4/9. We also know that P(X=x) = (1/3)*(2/3)^x for x=0,1,2,3. So we just need to find P(X=x | X>=2) for Show more…
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