A coin is tossed two times and if X denotes the number of heads obtained, find (i) E(x) (ii) E(X2) (iii) V(X) (iv) E(2X2 + 3x – 5)

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A coin is tossed two times and if X denotes the number of heads obtained, find (i) E(x) (ii) E(X2) (iii) V(X) (iv) E(2X2 + 3x &#8211; 5)

Suresh Chand · Accepted Answer

When a coin is tossed twice, there are four possible outcomes: HH, HT, TH, and TT. Let X denote the number of heads obtained, so X can take on the values 0, 1, or 2. The probability of getting 0, 1, or 2 heads is: P(X=0) = P(TT) = 1/4 P(X=1) = P(HT or TH) = 2/4 = 1/2 P(X=2) = P(HH) = 1/4 (i) The expected value of X is: E(X) = 0 x P(X=0) + 1 x P(X=1) + 2 x P(X=2) = 0 x 1/4 + 1 x 1/2 + 2 x 1/4 = 1 Therefore, E(X) = 1. (ii) The expected value of X^2 is: E(X^2) = 0^2 x P(X=0) + 1^2 x P(X=1) + 2^2 x P(X=2) = 0^2 x 1/4 + 1^2 x 1/2 + 2^2 x 1/4 = 1 Therefore, E(X^2) = 1. (iii) The variance of X is: V(X) = E(X^2) - [E(X)]^2 = 1 - 1^2 = 0 Therefore, V(X) = 0. (iv) The expected value of 2X^2 + 3X - 5 is: E(2X^2 + 3X - 5) = 2E(X^2) + 3E(X) - 5 = 2(1) + 3(1) - 5 = 0 Therefore, E(2X^2 + 3X - 5) = 0.

A coin is tossed two times and if X denotes the number of heads obtained, find (i) E(x) (ii) E(X²) (iii) V(X) (iv) E(2X² + 3x – 5)

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A coin is tossed two times and if X denotes the number of heads obtained, find (i) E(x) (ii) E(X2) (iii) V(X) (iv) E(2X2 + 3x – 5)

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A coin is tossed two times and if X denotes the number of heads obtained, find (i) E(x) (ii) E(X²) (iii) V(X) (iv) E(2X² + 3x – 5)