Consider a random variable X with the following probability function: 0.3 if x = 0, 0.2 if æ = 1, f(x) = 0.5 if x = 2, %3D 0 otherwise. (a) Find V(X) using the formula V(X) = E[(X – E(X))*]. (b) Using the result of (a) only, find V(3X + 3).

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A: Given: x P(x) 0 0.05 1 0.13 2 0.31 3 0.3 4 0.07

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A:

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A: Given,  P(m1) = 0.45 So, P(m0) = 1-0.45 = 0.55 P(B0|m0) = 0.40, So, P(B1|m0) = 1-0.4 =0.60 P(B0|m1)…

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A: Problem can be solved using concept of conditional probability (Bayes theorem).

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A: Given,E(X)=2Var(X)=4we know that…

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A: PDf of random variable X is:

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Q: Let X denote a random variable that takes on any of the values -1, 0, 1 with respective…

A: P(X = −1) = 0.2,P(X = 0) = 0.5,P(X = 1) = 0.3.

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A: There are five equally likely simple events,  so each event has probability .

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A: given p.d.f  p(x)=115(2x+1)   x = 1,2,3

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A:

Q: Given that X and Y are independent random variables and E(x) = 4, E(Y) = 7, Var(x) = 3, Var(Y) = 6;…

A: x and y are independent events. E(X) = 4 E(Y) = 7 Var(X) = 3 Var(Y) = 6   Var(X-6) = ?

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A:

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A: It is given that E(X)=2 and E(Y)=3.

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A: Here, we have

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A: Solution

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