This Exam P sample reference tests Joint Distributions. Revenue is the product of the two jointly distributed quantities. Summing that product against the six probability masses gives 140/9, so choice A is correct.
Let R denote the daily revenue. The supplied probability rule produces mass numerators 2, 4, 6 in the first row and 4, 2, 0 in the second row, all over 18.
R=XY
18pX,Y(x,y)=12102412421460
Model
Model
For a function of two discrete variables, multiply its value at each support point by the joint mass at that point.
E[R]=E[XY]=x∑y∑xypX,Y(x,y)
Compute
Compute
Evaluate the six weighted products and combine their numerators.
E[XY]=1820+48+84+80+48+0
E[XY]=18280=9140=15.555…
Answer
Answer
The expected revenue is 140/9.
9140(A)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThis replaces E[XY] by E[X]E[Y]. The marginals give E[X]=4/3 and E[Y]=12, whose product is 16=144/9, but X and Y are not independent.
CThis results if the first contribution is doubled by treating its x-coordinate as 2: replacing 20 by 40 changes the numerator from 280 to 300 and gives 150/9.
DThis results from copying the 80 numerator of the (2,10) contribution into the next cell instead of its correct 48; the total becomes 312/18=156/9.
EThis assigns equal weight to all six support points. Their unweighted revenue average is (10+12+14+20+24+28)/6=18, but the stated joint masses are unequal.
Original practice · fully worked
Original variant: expected data-transmission load
A field sensor chooses compression level C=1 with probability 0.60 and C=2 with probability 0.40. Conditional on C=1, the transmitted block count B is 3 with probability 0.75 and 5 otherwise. Conditional on C=2, B is 3 with probability 0.25 and 5 otherwise. Transmission load is L=CB. Calculate E[L].
A 5.30
B 5.46
C 5.70
D 6.25
E 10.00
Variant answer in brief
Conditioning on compression level gives mean loads 3.5 and 9.0. Weighting them by 0.60 and 0.40 gives 5.70, so choice C is correct.
Setup
Setup
Compute the conditional mean block count under each compression level.
E[B∣C=1]=3(0.75)+5(0.25)=3.5
E[B∣C=2]=3(0.25)+5(0.75)=4.5
Model
Model
Apply conditional expectation to the product L=CB.
E[L]=c=1∑2Pr(C=c)cE[B∣C=c]
Compute
Compute
Weight the two conditional load means by their level probabilities.
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