Independent solution

How to solve this Binomial Distribution question

Setup

Setup

Let R and C count the two treatment types across the group.

R∼Binomial⁡(15,0.9)R\sim\operatorname{Binomial}(15,0.9)
C∼Binomial⁡(15,0.4)C\sim\operatorname{Binomial}(15,0.4)
R⊥CR\perp C

Model

Model

Compute each count variance and express the aggregate payment.

Var⁡(R)=15(0.9)(0.1)=1.35\operatorname{Var}(R)=15(0.9)(0.1)=1.35
Var⁡(C)=15(0.4)(0.6)=3.6\operatorname{Var}(C)=15(0.4)(0.6)=3.6
T=2R+3CT=2R+3C

Compute

Compute

Independence removes the covariance term; scale variances by squared payment coefficients.

Var⁡(T)=22(1.35)+32(3.6)\operatorname{Var}(T)=2^2(1.35)+3^2(3.6)
Var⁡(T)=5.4+32.4=37.8\operatorname{Var}(T)=5.4+32.4=37.8

Answer

Answer

The total payment variance is 37.8.

37.8(B)\boxed{37.8\quad\text{(B)}}