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Suppose X is a random variable taking on possible values 1,2,3 with respective probabilities.4, .5, and .1. Y is a random var

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Solution:- from the data, Fonen X 1 3 PCz 0.4 3 PCY) O 3 O-3 O4 Sa Pobability that (x* y-u) PCXY4P (x= 2, yez) =P 60-5). PC=DStandard doviation ue Knouw dllat dtandasd deviatoo Varianca .: Variante Cx)- E C)-EC € (x2) ExtpCa 10 (x0.4) +(2x 05)+ (3x oGpected alue of Y E CY) EY PCY 3.1 + 11 .. E C)- 3.1T e Standand deuiadion at y Knous that S.D Vasiance ue SD CY)V Vaor Cy) Nf Expectod Vaue ef xy E CxY) E (x) ECY) 11 = 17 x 3- 1 5. 27 E (x *y)- 5.27 Expecteolvalue f 2x-3 y; E C2x-3Y)- 2(K)- 3E(Y) 1Please post the other questions separately. As per our answering guidelines we are supposed to answer just one question or four sub parts of the same question. Thank you.

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