This Exam P sample reference tests Multinomial Distribution. The two category counts have expectations 0.4 and 0.2. Applying the two loss amounts linearly gives expected total loss 2(0.4)+20(0.2)=4.80, which matches choice E.
How to solve this Multinomial Distribution question
Setup
Setup
Let I and J count the two nonzero outcome categories among the two trials. Their joint distribution is multinomial, so each count has expectation equal to the trial count times its category probability.
E[I]=2(0.20)=0.40
E[J]=2(0.10)=0.20
Model
Model
Express the total loss as a linear combination of the category counts.
L=2I+20J
E[L]=2E[I]+20E[J]
Compute
Compute
Substitute the two expected counts. No independence between I and J is needed for this expectation.
E[L]=2(0.40)+20(0.20)
E[L]=0.80+4.00=4.80
Answer
Answer
The expected total loss is 4.80.
E[L]=4.80(E)
Continue without hunting through PDFs
All 718 Exam P sample solutions in syllabus order
The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.60 is the expected number of nonzero-category outcomes, 2(0.20+0.10), rather than the expected monetary loss.
BThe value 1.00 is the sum of the probabilities in the multinomial distribution. Normalization alone does not weight outcomes by their losses.
CThe value 1.44 can arise by treating every nonzero outcome as a loss of two, obtaining the incorrect proxy 1.20, and then squaring it; expectation is linear and the second category has loss 20.
DThe value 3.92 omits the mixed outcome having one count of each type. Its probability-weighted contribution is 0.04(22)=0.88, and 4.80-0.88=3.92.
Original practice · fully worked
Original variant: instrument repair tickets
A laboratory receives four instrument-repair tickets. Each ticket independently requires no replacement part with probability 0.60, a sensor costing 5 credits with probability 0.30, or a rush controller costing 20 credits with probability 0.10. Calculate the expected total replacement-part cost for the four tickets.
A 1.60
B 3.50
C 6.00
D 8.00
E 14.00
Variant answer in brief
The expected sensor and rush-controller counts are 1.2 and 0.4. Their expected cost contributions are 6 and 8 credits, respectively, for a total of 14 credits and choice E.
Setup
Setup
Let S and R be the numbers of sensor and rush-controller replacements in the four tickets.
E[S]=4(0.30)=1.20
E[R]=4(0.10)=0.40
Model
Model
Write total parts cost as the corresponding linear combination of these counts.
C=5S+20R
E[C]=5E[S]+20E[R]
Compute
Compute
Evaluate the contribution from each replacement category.
E[C]=5(1.20)+20(0.40)
E[C]=6.00+8.00=14.00
Answer
Answer
The four tickets generate an expected replacement-part cost of 14 credits.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
