Independent solution

How to solve this Multinomial Distribution question

Setup

Setup

Let X, Y, and Z be the counts in the three categories across four independent trials, giving the stated multinomial model.

(X,Y,Z)∼Multinomial⁡(4;0.50,0.30,0.20)(X,Y,Z)\sim\operatorname{Multinomial}(4;0.50,0.30,0.20)

Model

Model

Enumerate every nonnegative count triple summing to four that also satisfies Z≥X+2.

Z≥X+2⟹(X,Y,Z)∈{(0,0,4),(1,0,3),(0,1,3),(0,2,2)}Z\ge X+2\Longrightarrow (X,Y,Z)\in\{(0,0,4),(1,0,3),(0,1,3),(0,2,2)\}

Compute

Compute

Evaluate the multinomial mass of each of the four feasible triples and add the disjoint probabilities to obtain 0.0488.

0.24+4(0.5)(0.2)3+4(0.3)(0.2)3+6(0.3)2(0.2)2=0.04880.2^4+4(0.5)(0.2)^3+4(0.3)(0.2)^3+6(0.3)^2(0.2)^2=0.0488

Answer

Answer

The requested probability rounds to 0.049, corresponding to choice D.

0.049(D)\boxed{0.049\quad\text{(D)}}