This Exam P sample reference tests Conditional Probability. Condition on the pooled alert by dividing the probability of exactly two affected subjects by the probability of at least one affected subject. The ratio is 0.00999966, so choice C is correct.
How to solve this Conditional Probability question
Setup
Setup
Let X count affected subjects among the three independent trials. The common affected probability is 0.01.
X∼Bin(3,0.01)
Model
Model
The observed pooled alert is equivalent to X being positive. Because the event X=2 is contained in that alert event, conditional probability reduces to a ratio.
Pr(X=2∣X≥1)=Pr(X≥1)Pr(X=2)
Compute
Compute
Use the binomial mass in the numerator and a complement in the denominator.
Pr(X=2)=(23)(0.01)2(0.99)=0.000297
Pr(X≥1)=1−(0.99)3=0.029701
0.0297010.000297=0.0099996633…
Answer
Answer
The requested conditional probability rounds to 0.0100.
0.0100(C)
Continue without hunting through PDFs
All 718 Exam P sample solutions in syllabus order
The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.000297 is the unconditional probability of exactly two affected subjects; it omits division by the observed-alert probability.
BThe value is obtained by retaining only one specified arrangement of two affected and one unaffected subject, thereby omitting the binomial factor of three.
DThe value 2(0.01)(0.99)=0.0198 is an exactly-one calculation for only two trials, so it uses neither the three-subject model nor the conditioning event.
EThe value 1-(0.99)³=0.029701 is the probability of the pooled alert itself, which is the denominator rather than the conditional result.
Original practice · fully worked
Original variant: mountain relay incident
A monitoring dashboard watches four independently operated mountain relays. Each relay has a 4% chance of missing its midnight heartbeat. The dashboard displays an incident whenever one or more heartbeats are absent. Given that an incident is displayed, calculate the probability that exactly two relays missed the heartbeat.
A 0.00885
B 0.00979
C 0.05873
D 0.06037
E 0.15065
Variant answer in brief
The exactly-two probability is 0.00884736 and the incident probability is 0.15065344. Their ratio is 0.0587266, which selects choice C.
Setup
Setup
Let M denote the number of relays that miss the heartbeat.
M∼Bin(4,0.04)
Model
Model
An incident means M is at least one. Exactly two misses automatically imply that an incident occurred.
Pr(M=2∣M≥1)=1−Pr(M=0)Pr(M=2)
Compute
Compute
Evaluate the two binomial quantities and form the conditional ratio.
Pr(M=2)=(24)(0.04)2(0.96)2=0.00884736
Pr(M≥1)=1−(0.96)4=0.15065344
0.150653440.00884736=0.0587265714…
Answer
Answer
The conditional probability is approximately 0.05873.
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.
