This Exam P sample reference tests Multinomial Distribution. This problem fixes the terminal observation as the fifth target-subtype case and assigns a multinomial composition to the first n-1 observations. The resulting joint mass function matches choice A.
How to solve this Multinomial Distribution question
Setup
Setup
Convert the nested classification rates into three mutually exclusive outcome probabilities: outside the broad category, inside it but outside the target subtype, and inside the target subtype.
p0=0.80
p1=0.20(1−0.08)=0.184
p2=0.20(0.08)=0.016
Model
Model
If the stopping time is n and the broad-category total is c, the terminal observation must be target subtype. Among the preceding n-1 observations, the three category counts are determined.
(n−c) of type 0,(c−5) of type 1,4 of type 2
N=n⟹observation n is type 2
Compute
Compute
Count the admissible orders of the first n-1 outcomes, multiply by their category probabilities, and append the terminal target-subtype probability.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThe factors 0.12 and 0.08 treat the conditional subtype rate 0.08 as a population-wide probability. The correct mutually exclusive probabilities are 0.184 and 0.016 after multiplying by the broad-category rate 0.20.
CThe coefficient n!/[(n-c)!(c-5)!5!] places all five target-subtype observations freely among all n positions. It fails to require the fifth one to occur at the terminal observation.
DThis coefficient fixes all five target-subtype positions. It omits the combinations that place the first four target cases among the first n−1 observations.
EIts first n-1 category counts use n-c-1 and c-4, so the broad-category total becomes c+1 after the terminal target case is appended. It also uses the unconverted conditional rates 0.12 and 0.08.
Original practice · fully worked
Original variant: equal counts in two categories
Four independent calibration results are classified as fast with probability 0.50, standard with probability 0.30, or deferred with probability 0.20. Calculate the probability that the number of fast results equals the number of standard results.
A 0.0016
B 0.0720
C 0.1350
D 0.2086
E 0.7914
Variant answer in brief
Equal fast and standard counts can be 0 and 0, 1 and 1, or 2 and 2. The corresponding multinomial probabilities are 0.0016, 0.0720, and 0.1350, totaling 0.2086 and choice D.
Setup
Setup
Let F, S, and D be the three category counts. If F equals S, the four results permit only three count patterns.
(F,S,D)∈{(0,0,4),(1,1,2),(2,2,0)}
Model
Model
Evaluate the multinomial probability of each disjoint pattern.
p004=(0.20)4=0.0016
p112=1!1!2!4!(0.50)(0.30)(0.20)2=0.0720
p220=2!2!4!(0.50)2(0.30)2=0.1350
Compute
Compute
Add the three mutually exclusive ways for the two counts to agree.
Pr(F=S)=0.0016+0.0720+0.1350=0.2086
Answer
Answer
The fast and standard counts agree with probability 0.2086.
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