This Exam P sample reference tests Binomial Distribution. For a binomial count with success probability 0.60, the ratio of the probability of three successes to that of two is (n-2)/2. Setting this ratio equal to five gives n=12, so choice D is correct.
Equate the simplified ratio to the stated multiple and solve for the integer number of trials.
2n−2=5
n=12
Answer
Answer
The season contains 12 games.
12(D)
Test the method on unseen problems
General Probability Practice Workbook
30 original general probability problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AAt n=6 the adjacent-probability ratio is (6-2)/2=2, not the required five.
BAt n=8 the adjacent-probability ratio is (8-2)/2=3. This results from matching the ratio to three instead of the required five.
CAt n=10 the ratio is (10-2)/2=4, an off-by-one result from replacing the required multiplier five by four.
EAt n=14 the ratio is (14-2)/2=6, an off-by-one result from replacing the required multiplier five by six.
Original practice · fully worked
Original variant: variability from adjacent Poisson counts
The number N of photon artifacts found on a scan is Poisson distributed with an unknown mean. The probability of exactly five artifacts is three halves the probability of exactly four artifacts. Calculate the standard deviation of N.
A 1.500
B 2.449
C 2.739
D 5.000
E 7.500
Answer and calculation
For a Poisson variable, P(N=5)/P(N=4)=λ/5. The reported ratio gives λ=7.5, so SD(N)=√(7.5)=2.7386 and choice C.
Setup
Setup
Write the two adjacent Poisson masses using the unknown positive mean λ.
Pr(N=k)=e−λk!λk
Model
Model
Divide the mass at five by the mass at four so that the common exponential factor and four powers cancel.
Pr(N=4)Pr(N=5)=e−λλ4/4!e−λλ5/5!=5λ
Compute
Compute
Recover the Poisson mean from the ratio, then use equality of its mean and variance.
5λ=23⟹λ=215=7.5
SD(N)=Var(N)=λ=7.5=2.738612788
Answer
Answer
The artifact-count standard deviation is approximately 2.739.