This Exam P sample reference tests Multinomial Distribution. Enumerating the four multinomial count patterns satisfying the inequality gives total probability 0.0488, which rounds to choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThe cells (0,0,4) and (0,1,3) have total probability 0.0016+0.0096=0.0112, approximately 0.012; this omits the other two feasible cells.
CThe two feasible cells with Y=0 have total probability 0.0016+0.0160=0.0176, approximately 0.018; this omits the cells with positive Y.
Original practice · fully worked
Original variant: one cell of a three-category allocation
Five independent packages are classified as Green, Amber, or Red. For each package the respective classification probabilities are 0.45, 0.35, and 0.20. Find the probability that the batch contains exactly two Green, one Amber, and two Red packages.
A 0.08505
B 0.07200
C 0.06000
D 0.04500
E 0.01800
Variant answer in brief
The multinomial coefficient is 5!/(2!1!2!)=30. Multiplying by the category powers gives 0.08505, choice A.
Setup
Setup
Model the Green, Amber, and Red package counts as a three-category multinomial vector over five independent packages.
(G,A,R)∼Multinomial(5;0.45,0.35,0.20)
Model
Model
For counts 2, 1, and 2, multiply the category probability powers by the multinomial arrangement count 5!/(2!1!2!).
Pr(2,1,2)=2!1!2!5!(0.45)2(0.35)(0.20)2
Compute
Compute
There are 30 arrangements, and their common probability product gives a total mass of 0.08505.
Pr(2,1,2)=30(0.2025)(0.35)(0.04)=0.08505
Answer
Answer
The probability of exactly two Green, one Amber, and two Red packages is 0.08505, which is choice A.
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