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.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.32 is 0.75⁴ = 0.316406 after rounding. It corresponds to needing more than four attempts, an off-by-one tail.
CThe value 0.44 is 1-0.75² = 0.4375 after rounding, the probability of success within the first two attempts.
DThe value 0.56 is 0.75² = 0.5625 after rounding. It requires only the first two attempts to fail.
EThe value 0.58 is 1-0.75³ = 0.578125, the complement: success occurs within the first three attempts.
Original practice · fully worked
Original variant: negative-binomial completion tail
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.
A 0.046656
B 0.311040
C 0.419904
D 0.455680
E 0.544320
Variant answer in brief
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
Setup
Translate the completion-time statement into a count statement for the first six readings.
{T3>6}={N6≤2}
Model
Model
The accepted-reading count in six independent trials is binomial.
N6∼Binomial(6,0.40)
Compute
Compute
Add the probability masses for zero, one, and two acceptances.
Pr(N6≤2)=k=0∑2(k6)(0.4)k(0.6)6−k
=0.046656+0.186624+0.311040=0.544320
Answer
Answer
The routine exceeds six readings with probability 0.544320.
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