Question

In an experiment, a random variable X can take values 0, 1, 2, 3. If P(X = 0) = P(X = 1), P(X = 2) = P(X = 3) and \(E\left(X^{2}\right)=2 E(X),\) then the value of P(X = 0) is

(1) \(\frac{1}{8}\)

(2) \(\frac{1}{4}\)

(3) \(\frac{1}{2}\)

(4) \(\frac{3}{8}\)

Answer 1

Correct option is: (4) \(\frac{3}{8}\)  

\(P(X=0)=P(x=1)=a\) and \(P(X=2)=P(X=3)=b\)

\(2 a+2 b=1\)

\(\Rightarrow a+b=\frac{1}{2}\)

\(E\left(X^{2}\right)=2 E(X)\)

\(=a \times 0^{2}+a \times 1^{2}+b X \times 2^{2}+b \times 3^{2}\)

\(=2(a \times 0+a \times 1+b \times 2+b \times 3)\)  

\(\Rightarrow a=3 b\)

\(4 b=\frac{1}{2} \Rightarrow b=\frac{1}{8}\)

\(\therefore a=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

\(P(X=0)=a=\frac{3}{8}\)