This Exam P sample reference tests Conditional Probability. Conditioning leaves count two with weight 2/3 and count three with weight 1/3. The corresponding threshold probabilities are 1/4 and 7/8, giving 11/24=0.4583 and choice D.
How to solve this Conditional Probability question
Setup
Setup
Let N be the event count. Once the stated count condition is imposed, only the two-count and three-count cases remain.
Pr(N=2∣N≥2)=0.2+0.10.2=32
Pr(N=3∣N≥2)=31
Model
Model
With two events, the threshold is reached only when both severities take their larger value. With three events, it is reached unless all three take their smaller value.
q2=(0.5)2=41
q3=1−(0.5)3=87
Compute
Compute
Average the two conditional threshold probabilities using the conditioned count weights.
Pr(S≥2000∣N≥2)=32(41)+31(87)
Pr(S≥2000∣N≥2)=2411=0.4583333333
Answer
Answer
The conditional probability rounds to 0.46.
0.46(D)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis calculates the unconditioned joint probability 0.2(1/4)+0.1(7/8)=0.1375 and rounds it to 0.14, but never divides by Pr(N≥2)=0.30.
BThis reports the two-event conditional threshold probability (0.5)²=0.25 and ignores the possible three-event case.
CThis keeps only the three-event contribution after conditioning, (1/3)(7/8)=7/24=0.2917, and omits the two-event contribution.
EThis reports 1-(0.5)³=0.875, the threshold probability given exactly three events, rather than mixing both allowed counts.
Original practice · fully worked
Original variant: heavy-kit dispatch inference
A warehouse dispatches 1, 2, or 4 replacement kits with probabilities 0.50, 0.30, and 0.20. Each kit independently weighs 1 kg with probability 0.75 or 3 kg with probability 0.25. Given that a dispatch weighs at least 6 kg, calculate the probability that it contains four kits.
A 0.1367
B 0.4000
C 0.6836
D 0.8794
E 0.9162
Variant answer in brief
A two-kit dispatch qualifies only when both kits are heavy, while a four-kit dispatch qualifies when at least one is heavy. Bayes' formula gives 175/199=0.8794, so choice D is correct.
Setup
Setup
Let H be the event that total dispatch weight is at least 6 kg. A one-kit dispatch cannot satisfy H.
Pr(H∣N=1)=0
Model
Model
Two kits require two heavy outcomes. Four kits already weigh 4 kg, so at least one heavy outcome is sufficient.
Pr(H∣N=2)=(41)2=161
Pr(H∣N=4)=1−(43)4=256175
Compute
Compute
Apply Bayes' formula to the two count values that can produce the observed weight.
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.
