For Binomial: E(X)=np=12⋅31=4.
Var(X)=npq=12⋅31⋅32=38.
E(X2)=Var(X)+(E(X))2=38+16=356.
E(X2)+E(X)=356+4=356+12=368.
In Binomial distribution with parameters n = 12 and p = 1 3, value of E(X^2) + E(X) is :
CUET UG 2022 — Mathematics Algebra
In Binomial distribution with parameters n=12 and p=31, value of E(X2)+E(X) is :
Held on 7 Aug 2022 · Verified 13 Jul 2026.
Options
- A
1
- B
38
- C
356
- D
368
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