This Exam P sample reference tests Normal Distribution. Split by zero-claim indicators. The certain positive-versus-zero case contributes 0.18, and the both-positive normal comparison contributes 0.04342, totaling 0.22342, choice D.
Separate the comparison into activity states. If service B is positive while service A is zero, B certainly exceeds A and contributes probability 0.18.
Pr(B>0,A=0)=0.30(0.60)=0.18
Model
Model
When both services are positive, their difference is normal. Its mean is negative 1,000 and its variance is the sum of the two independent variances.
B−A∣(A>0,B>0)∼N(−1000,20002+20002)
Compute
Compute
Weight the positive-difference probability by the probability that both services are active, then add the certain positive-versus-zero case. The result is approximately 0.223421.
Pr(B>A)=0.18+0.12Φ(8,000,000−1000)=0.223421
Answer
Answer
The unconditional probability that B exceeds A is approximately 0.223, selecting choice D.
0.223(D)
Test the method on unseen problems
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.180 includes only the case where B is positive and A is zero. It omits the possibility that B exceeds A when both are positive.
Original practice · fully worked
Original variant: conditional comparison of two zero-inflated Gaussian totals
Service A is inactive with probability 0.40; when active its total is normal with mean 5 and standard deviation 1. Service B is inactive with probability 0.60; when active its total is normal with mean 7 and standard deviation 1.5. The services are independent. Given that at least one is active, find P(B's total exceeds A's total).
A 0.21053
B 0.31579
C 0.48412
D 0.52632
E 0.76000
Answer and calculation
The unconditional comparison numerator is 0.367929 and at-least-one activity has probability 0.76. Their ratio is 0.48412, choice C.
Setup
Setup
Build the unconditional numerator by separating the case where only B is active from the case where both services are active. In the latter case, compare the two independent normal totals through their difference.
Pr(B>A)=0.40(0.40)+0.40(0.60)Φ(2/3.25)=0.367929
Model
Model
The conditioning event excludes only the state in which both services are inactive, so its probability is 0.76.
Pr(A>0 or B>0)=1−0.40(0.60)=0.76
Compute
Compute
The unconditional favorable probability is approximately 0.367929. Dividing by 0.76 gives approximately 0.484117.
Pr(B>A∣A>0 or B>0)=0.367929/0.76=0.484117
Answer
Answer
After conditioning on at least one active service, the comparison probability is approximately 0.48412, selecting choice C.
