Worked solution · Exam P
This Exam P sample reference tests Negative Binomial Distribution. This is a negative-binomial waiting-time event for the second failure. Exactly one of the first three trials must fail and the fourth must fail, giving 3(0.02)²(0.98)²=0.00115248 and choice B.
Independent solution
Setup
Let p be the failure probability on one independent trial and q the success probability.
Model
For the second failure to occur on trial four, the first three trials must contain exactly one failure and trial four must be a failure.
Compute
Substitute the one-trial probabilities.
Answer
The probability rounds to 0.00115.
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
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
Original practice · fully worked
Two teams play a best-of-five championship series. Team A wins each game independently with probability 0.60. Calculate the probability that the series ends after exactly four games.
The series can end in four games with either team winning. Adding the disjoint probabilities that A wins 3–1 and that B wins 3–1 gives 0.2592+0.1152=0.3744, so choice D.
Setup
The series ends on game four only if the eventual winner has exactly two wins in the first three games and wins game four.
Model
Compute the two disjoint cases: Team A clinches in game four or Team B clinches in game four.
Compute
Add the two ways the series can terminate after four games.
Answer
The probability that the series lasts exactly four games is 0.3744.
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