The probability distribution of a random variable X is x 0 1 2 3 P(X…

$E(x^2)=0^2×\frac{1}{4}+(1)^2×\frac{1}{8}+(2)^2×\frac{1}{8}+(3)^2×\frac{1}{2}$

$=0+\frac{1}{8}+\frac{4}{8}+\frac{9}{2×4}⇒\frac{1+4+36}{8}=\frac{41}{8}$

$E(x)=0×\frac{1}{8}+2×\frac{1}{8}+3×\frac{1}{2}=\frac{1}{8}+\frac{2}{8}+\frac{3}{2×4}$

$\frac{1+2+12}{8}=\frac{15}{8}$

Variance(X) = $E[x^2]-E[x]^2=\frac{41}{8}-(\frac{15}{8})^2$

$=\frac{41}{8×8}-\frac{225}{64}=\frac{328-225}{64}$

$=\frac{103}{64}$

Option 3rd is correct.