This Exam P sample reference tests Sums of Independent Random Variables. A total below two consists of no losses, one hail loss, or one theft loss. The three independent-event probabilities sum to 0.1726357, so choice A is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThis collapses each year to an indicator for exactly one loss, with probability 0.2(0.9)+0.1(0.8)=0.26, and then computes P(Binomial(10,0.26)≤1)=0.222245, losing years with two losses.
CThis uses the same incorrect one-loss-per-year model as choice B and then takes its complement, 1-0.222245=0.777755.
DThis approximates the total by Poisson(3) and then takes the complement of the desired small-count event: 1-4 exp(−3)=0.800852.
EThis takes the complement of the exact result, 1-0.1726357=0.8273643, thereby answering for at least two losses.
Original practice · fully worked
Original variant: infer an alarm rate
Two alarm channels are observed for six shifts. Channel A triggers independently on each shift with probability 0.10. Channel B is independent of A and across shifts, but its trigger probability q is unknown. The probability that neither channel triggers during all six shifts is 0.0945851. Calculate the expected total number of triggers from both channels over the six shifts.
A 0.25
B 0.35
C 0.60
D 1.50
E 2.10
Variant answer in brief
The no-trigger probability is (0.9)⁶(1-q)⁶=0.0945851, so 0.9(1-q)=0.675 and q=0.25. Linearity gives expected total 6(0.10+0.25)=2.10, selecting E.
Setup
Setup
Write the six-shift probability of no trigger from either independent channel.
0.0945850803=0.96(1−q)6
Model
Model
Take the sixth root to recover the per-shift no-trigger factor.
0.9(1−q)=0.09458508031/6=0.675
Compute
Compute
Solve for q and add the expected trigger counts from the two channels.
q=0.25
E[NA+NB]=6(0.10)+6(0.25)=2.10
Answer
Answer
The two channels produce 2.10 triggers on average.
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