If X is a random variable which can take values 0,1,2,3 such that \( \mathrm{E\,(X^2)=2E\,(X)} \).
If P (X = 0) = P (X = 1) = m & P(X = 2) = 2m, then which of the following are TRUE?
A. \( \mathrm{P(X=1)}=\displaystyle\frac{1}{13} \)
B. \( \mathrm{P(X=3)}=\displaystyle\frac{1}{13} \)
C. \( \mathrm{E(X)}=\displaystyle\frac{18}{13} \)
D. \( \mathrm{P(X=2)}=\displaystyle\frac{6}{13} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B, C and D only
The probabilities add up to 1, so
\( \displaystyle P(X=3)=1-(m+m+2m)=1-4m \)
\( \displaystyle E(X)=0(m)+1(m)+2(2m)+3(1-4m) \)
\( \displaystyle =3-7m \)
\( \displaystyle E(X^2)=0(m)+1(m)+4(2m)+9(1-4m) \)
\( \displaystyle =9-27m \)
Using \( E(X^2)=2E(X) \):
\( \displaystyle 9-27m=6-14m \)
\( \displaystyle 13m=3\ \Rightarrow\ m=\frac{3}{13} \)
\( \displaystyle P(X=2)=2m=\frac{6}{13} \)
\( \displaystyle P(X=3)=1-\frac{12}{13}=\frac{1}{13} \)
\( \displaystyle E(X)=3-\frac{21}{13}=\frac{18}{13} \)
So P(X = 1) = 3/13, P(X = 2) = 6/13, P(X = 3) = 1/13 and E(X) = 18/13.
Checking the statements
(A) \( \mathrm{P(X=1)}=\frac{1}{13} \) — Incorrect. \( P(X=1)=m=\frac{3}{13} \).
(B) \( \mathrm{P(X=3)}=\frac{1}{13} \) — Correct. \( 1-4\times\frac{3}{13}=\frac{1}{13} \).
(C) \( \mathrm{E(X)}=\frac{18}{13} \) — Correct. \( 3-7\times\frac{3}{13}=\frac{18}{13} \).
(D) \( \mathrm{P(X=2)}=\frac{6}{13} \) — Correct. \( 2m=\frac{6}{13} \).