Worked solution · Exam P
This Exam P sample reference tests Geometric Distribution. The condition fixes the first roll as a six, so it cannot also be the first five. Beginning with roll two, the wait for a five is geometric with mean 6; adding the observed first roll gives a conditional expected total of 7 and choice E.
Independent solution
Setup
Interpret the condition as fixing the outcome of the first roll.
Model
After that observed roll, independent future rolls restart the wait for the first five. Let G count future rolls through the first five.
Compute
Use the trial-count convention for the geometric mean and add the completed first roll.
Answer
The conditional expected number of rolls is seven.
Test the method on unseen problems
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.
Error diagnosis
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Original practice · fully worked
An access token is assigned to verification route A with probability 0.60 and route B with probability 0.40. Conditional on its assigned route, repeated attempts are independent and use that same route until one is accepted. Each attempt is accepted with probability 0.50 on route A and 0.25 on route B. Given that the first attempt is rejected, calculate the expected total number of attempts through the first acceptance.
The first rejection makes the two route posterior probabilities equal: each contributes probability 0.30 to the rejection event. The expected remaining attempts are therefore the average of geometric means 2 and 4, and adding the rejected attempt gives 4.0 and choice D.
Setup
Weight the first-rejection likelihood by each route's prior probability.
Model
Normalize the equal contributions to update the hidden route after the rejection.
Compute
Future attempts have geometric means two and four under the two routes. Average them with the posterior weights, then add the completed rejection.
Answer
The conditional expected total is four attempts.
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