we will consider a random variable whose pdf is the form of f(x) = Cx™(1-x) where x E (0, 1) and m is a non-negative integer. (a) Consider the special situation of m = 1. Find the value of C such that the pdf is a proper pdf. (b) Argue that for a general m, the pdf must be f (x) = (m + 1)(m + 2)x™(1 – x) for x E (0, 1).

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A: Given,X≈b(12,23 )E(x)=npE(x)=12×23 E(x)=8