This Exam P sample reference tests Conditional Probability. This problem recovers a group proportion from a joint probability and a within-group conditional probability. Dividing 0.30 by 0.50 gives a male proportion of 0.60, so the complementary female proportion is 0.40 and choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AMultiplying 0.30 by 0.50 gives 0.15. The intersection is already supplied; the conditional relation must be inverted to recover the group proportion.
CThe value 0.50 is the accident probability within the male group, not the female marginal probability.
DThe value 0.60 is the recovered male proportion. The question asks for its complement.
EThe value 0.85 is the complement of the erroneous product 0.15, carrying the initial multiplication error into the final step.
Original practice · fully worked
Original variant: encrypted records without signatures
Seventy-five percent of archived records are encrypted. Among encrypted records, 40% carry a digital signature. Calculate the probability that a randomly selected record is encrypted but unsigned.
A 0.250
B 0.300
C 0.400
D 0.450
E 0.600
Variant answer in brief
An encrypted record is unsigned with conditional probability 0.60. Multiplying by the encrypted share 0.75 gives 0.45 and choice D.
Setup
Setup
Complement the signature rate inside the encrypted group.
Pr(Sc∣E)=1−0.40=0.60
Model
Model
Convert the within-group rate to a joint probability.
Pr(E∩Sc)=Pr(E)Pr(Sc∣E)
Compute
Compute
Multiply the group share by its unsigned fraction.
Pr(E∩Sc)=0.75(0.60)=0.45
Answer
Answer
Forty-five percent of all records are both encrypted and unsigned.
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