Question

(3) (6 points) Suppose that X is a random variable having possible values and 1. Suppose that P(X = 0) = 3P(X = 1). -2 (a) Find E(X) (b) What is P(X=)? (c) Find Var(X).

Answer 1

SOlution Let be random vanable takes Value -2 and uppase PCX-2) 3. p(x=) you have not replaced have Jau yer O by-2 given Plxzo) 3P Cx1) Suppose Pix=1): k. P(-2)3k detof pababiliy ΣΡΧ) thin Kt 3 k- pCas-2) 3/4 and Ca) E (x) = E (x) E (x) - 5 ka 4 Since rom 3/4 PCx-2) 3K- CC Var Cx) ? Var (x) pczza) - (बजाड शा हे फो 22 3+ 4 25 13 Var x) 27/16

condihons by as take alue and -2 And PCK- O3PCXi) thin because PCx-0)-0 not takes Yaluc o Hence in given Condiwon. as but Px0+ PCKl) + P (x2) P(X2) pCxz-2) = (a) E CX)- (-2) Σχρ(κ) - 2 P(X= -2) : 1 Vorr (x) (-2)2 Varr x)