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If X is a random variable which can take values 0,1,2,3 such that E(X2)=2E(X)E(X^2) = 2E(X).

If P(X=0)=P(X=1)=m & P(X=2)=2mP(X=0) = P(X=1) = m\ \&\ P(X=2) = 2m, then which of the following are TRUE?

A. P(X=1)=113P(X=1) = \dfrac{1}{13}

B. P(X=3)=113P(X=3) = \dfrac{1}{13}

C. E(X)=1813E(X) = \dfrac{18}{13}

D. P(X=2)=613P(X=2) = \dfrac{6}{13}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

P(X=3)=1−4mP(X=3)=1-4m. Then E(X)=3−7mE(X)=3-7m and E(X2)=9−27mE(X^2)=9-27m.

Setting 9−27m=2(3−7m)9-27m=2(3-7m) gives m=313m=\dfrac{3}{13}.

So P(X=1)=313P(X=1)=\dfrac{3}{13} (A false), P(X=3)=113P(X=3)=\dfrac{1}{13}, P(X=2)=613P(X=2)=\dfrac{6}{13} and E(X)=1813E(X)=\dfrac{18}{13}.

Hence B, C and D are true.

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