Worked solution · Exam P
This Exam P sample reference tests Geometric Distribution. Needing more than three attempts means that each of the first three attempts fails. Independence gives 0.75³ = 0.421875, which rounds to choice B.
Independent solution
Setup
Let T be the attempt number of the first success. A single attempt fails with probability 0.75.
Model
The event T>3 is exactly the event that attempts one through three are all failures.
Compute
Multiply the three independent failure probabilities.
Answer
The listed value is 0.42.
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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.
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Original practice · fully worked
A calibration routine requires three accepted readings. Each reading is independently accepted with probability 0.40. Calculate the probability that the routine needs more than six readings to collect its third accepted reading.
More than six readings are needed exactly when the first six contain at most two acceptances. The Binomial(6,0.4) probabilities for zero, one, and two successes sum to 0.544320, choice E.
Setup
Translate the completion-time statement into a count statement for the first six readings.
Model
The accepted-reading count in six independent trials is binomial.
Compute
Add the probability masses for zero, one, and two acceptances.
Answer
The routine exceeds six readings with probability 0.544320.
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