This Exam P sample reference tests Geometric Distribution. If p is the per-trial event probability, the supplied second-trial probability gives (1-p)p=0.16. The stated parameter restriction selects p=0.2, after which (0.8)³(0.2)=0.1024 and choice D is correct.
Let p be the constant event probability on one trial and q=1-p. Let T denote the trial on which the first event occurs.
Pr(T=k)=qk−1p
Model
Model
Use the supplied second-trial probability to identify p, retaining the parameter restriction when choosing between the two quadratic roots.
qp=(1−p)p=0.16
p2−p+0.16=0
Compute
Compute
Solve the quadratic, select the admissible root, and evaluate the geometric mass at four.
p=21±1−0.64∈{0.2,0.8}
p<0.5⟹p=0.2,q=0.8
Pr(T=4)=q3p=(0.8)3(0.2)=0.1024
Answer
Answer
The probability that the first event is on trial four is 0.1024.
0.1024(D)
Test the method on unseen problems
Probability Univariate Distributions Practice Workbook
30 original univariate random variables 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.
AChoosing the excluded root p=0.8 gives q=0.2 and q³p=(0.2)³(0.8)=0.0064. The parameter restriction is what rules out this root.
BThe value 0.0256=(0.16)² squares the given second-trial probability. First-event masses at different times are not independent events that can be multiplied this way.
CUsing 1-3p=0.4 as an approximation to the probability of no event in the first three trials gives 0.4(0.2)=0.0800. Independence requires q³=0.8³=0.512 instead.
EThe value 0.2000 is the per-trial event probability p. A first event on trial four also requires three preceding non-events, contributing the factor q³.
Original practice · fully worked
Original variant: retry-log probability ratio
A server repeats independent transmission attempts until the first acknowledgement, with the same acknowledgement probability on every attempt. The probability that the first acknowledgement occurs on attempt five is 0.343 times the probability that it occurs on attempt two. Given that attempt one failed, calculate the probability of an acknowledgement by the end of attempt four.
A 0.300
B 0.343
C 0.490
D 0.657
E 0.700
Answer and calculation
The ratio of the two first-acknowledgement masses is q³=0.343, so the failure probability is q=0.7. Conditional on the first failure, success within the next three attempts has probability 1-q³=0.657, so choice D is correct.
Setup
Setup
Let p be the acknowledgement probability and q=1-p the failure probability on each independent attempt.
Pr(T=k)=qk−1p
Model
Model
Form the ratio of the two supplied first-acknowledgement probabilities; the common factor p cancels.
Pr(T=2)Pr(T=5)=qpq4p=q3=0.343
Compute
Compute
Infer q and complement the probability that all three remaining attempts through attempt four fail.
q=30.343=0.7,p=0.3
Pr(T≤4∣T>1)=1−q3=1−0.343=0.657
Answer
Answer
Given the first failed attempt, the chance of an acknowledgement by attempt four is 0.657.
