A small discrete distribution

Problem

Let X∼Binomial⁡(3,1/2)X\sim\operatorname{Binomial}(3,1/2). List its pmf for x=0,1,2,3x=0,1,2,3, then compute P(X≥2)P(X\ge2), E[X]E[X], and Var⁡(X)\operatorname{Var}(X).

Reveal answer or reference solution
P(X=x)=(3x)2−3,P(X=x)=\binom3x2^{-3},

so the probabilities for x=0,1,2,3x=0,1,2,3 are (1,3,3,1)/8(1,3,3,1)/8. Therefore

P(X≥2)=38+18=12P(X\ge2)=\frac38+\frac18=\frac12

For a binomial variable, E[X]=np=3/2E[X]=np=3/2 and Var⁡(X)=np(1−p)=3/4\operatorname{Var}(X)=np(1-p)=3/4.

Outcome

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