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In Binomial distribution with parameters n=12n = 12 and p=13p = \frac{1}{3}, value of E(X2)+E(X)E(X^2) + E(X) is :

Solution

✅ Correct Option: 4

For Binomial: E(X)=np=12⋅13=4E(X) = np = 12 \cdot \frac{1}{3} = 4.

Var(X)=npq=12⋅13⋅23=83\text{Var}(X) = npq = 12 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3}.

E(X2)=Var(X)+(E(X))2=83+16=563E(X^2) = \text{Var}(X) + (E(X))^2 = \frac{8}{3} + 16 = \frac{56}{3}.

E(X2)+E(X)=563+4=56+123=683E(X^2) + E(X) = \frac{56}{3} + 4 = \frac{56 + 12}{3} = \frac{68}{3}.

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