This Exam P sample reference tests Binomial Distribution. Taking the ratio of the adjacent Binomial(100,p) masses cancels their large common factors. The condition becomes (98/3)p/(1-p)=2, giving p=3/52=0.0576923 and choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.040 results from setting the expected count to four by combining the referenced count two with the factor of two, then dividing by 100. The statement compares probability masses, not the mean count.
BStarting from that erroneous 0.040 rate and then reporting its odds gives 0.040/0.960=0.04167, or 0.042. Neither step uses the adjacent-mass ratio.
DA Poisson shortcut gives P(X=3)/P(X=2)=λ/3=2, hence λ=6 and p=λ/100=0.060. That is only an approximation; the exact binomial finite-n factor gives 0.057692.
EAt p=0.072, the exact adjacent-mass ratio is (98/3)(0.072/0.928)=2.534, not 2. This choice reflects an incorrect binomial-coefficient or trial-count factor.
Original practice · fully worked
Original variant: count spread from a conditional audit
A review batch contains eight independent records, each accepted with the same unknown probability p. Let X be the accepted-record count. Conditional on X being either 2 or 3, an audit reports probability 0.60 that X equals 3. Calculate the standard deviation of X.
A 0.4286
B 0.6000
C 1.3997
D 1.9592
E 3.4286
Variant answer in brief
The conditional audit implies adjacent-mass ratio P(X=3)/P(X=2)=0.60/0.40=3/2. For Binomial(8,p) that ratio is 2p/(1-p), giving p=3/7 and standard deviation √(96/49)=1.3997, choice C.
Setup
Setup
Model the accepted-record count as binomial with eight trials.
X∼Binomial(8,p)
Model
Model
Convert the conditional audit percentage into odds between the two adjacent counts.
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.