This Exam P sample reference tests Conditional Probability. The given no-theft probabilities determine a four-cell table: the probability that both households have a theft is 0.10, while the conditioning household has a theft with probability 0.20. Their ratio is 0.50, choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.20(0.30)=0.060 treats the chance of at least one theft as though it were B's theft probability and also imposes unsupported independence.
BThe value 0.100 is the joint probability of thefts at both households; it is the numerator before conditioning on B's theft.
CThe value 0.200 is A's marginal theft probability and ignores the information about what happened at household B.
EThe ratio 0.70/0.80=0.875 is a conditional no-theft probability, Pr(T_Bᶜ given T_Aᶜ), for the complementary events.
Original practice · fully worked
Original variant: alpine telemetry links
A control center receives telemetry through a north link and a south link. Each link is available at the audit time with probability 0.85, and both are available with probability 0.75. If the south link is unavailable, calculate the probability that the north link is also unavailable.
A 0.050
B 0.150
C 0.250
D 0.333
E 0.882
Variant answer in brief
Inclusion–exclusion gives a 0.05 probability that both links are unavailable. The south link is unavailable with probability 0.15, so the conditional probability is 0.05/0.15=1/3, choice D.
Setup
Setup
Let U_N and U_S denote unavailability of the north and south links.
Pr(UN)=Pr(US)=0.15
Pr(UNc∩USc)=0.75
Model
Model
Use the availability information to find the lower-right cell of the two-link status table.
Pr(UN∩US)=1−0.85−0.85+0.75
Compute
Compute
Normalize the both-unavailable cell by the south-unavailable marginal probability.
Pr(UN∩US)=0.05
Pr(UN∣US)=0.150.05=31
Answer
Answer
The required conditional probability is approximately 0.333.
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