This Exam P sample reference tests Binomial Distribution. Revenue starts at 1,050. A single 100 refund occurs only if all 21 ticket holders arrive, an event of probability 0.98²¹, giving expected revenue about 985 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 955 implies expected refund 95 and therefore penalty probability 0.95, not the all-arrive probability about 0.6543.
BThe value 962 implies penalty probability 0.88. The refund event is all 21 arriving, not at least one no-show.
CThe value 967 implies penalty probability 0.83, which is not obtained from the 21 independent arrival events.
DThe value 976 implies penalty probability 0.74; no standard rounding of 0.98 to the 21st power produces that value.
Original practice · fully worked
Original variant: expected reserve after a multi-failure penalty
A server operator starts a week with a 600-dollar maintenance reserve. Five independent backup disks each fail with probability 0.10. If at least two disks fail, a one-time 150-dollar emergency charge is paid. Find the expected reserve remaining.
A 565.00 dollars
B 575.78 dollars
C 581.46 dollars
E 587.78 dollars
D 592.50 dollars
Variant answer in brief
The chance of at least two failures is 1-0.9⁵-5(0.1)0.9⁴=0.08146. The expected emergency charge is 12.219 dollars, leaving 587.781.
Setup
Setup
Let N be the number of failures among five independent disks. The reserve loses 150 dollars exactly when N is at least 2.
N∼Bin(5,0.1)
R=600−150I{N≥2}
Model
Model
Find the binomial upper tail by subtracting the zero- and one-failure probabilities from one.
P(N≥2)=1−0.95−5(0.1)(0.9)4=0.08146
Compute
Compute
The emergency-charge probability is 0.08146, so its expected cost is 12.219 dollars and the expected remaining reserve is 587.781.
E[R]=600−150(0.08146)=587.781
Answer
Answer
Expected remaining reserve is 587.78 dollars, corresponding to choice E.
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