This Exam P sample reference tests Uniform Distribution. This problem translates payment conditions back to ranges of a uniform loss. The target intersection has width 100 inside a conditioning interval of width 800, giving probability 100/800=0.125 and choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.100 is the unconditional probability of the 100-unit target interval within the full 1000-unit support; it omits conditioning.
CRestricting to positive payments as well as payments below 400 gives 100/400=0.250. The stated condition also includes zero payments.
DThe value 300/400=0.750 computes the complementary range among positive payments below 400, both changing the condition and reversing the target.
EThe value 700/800=0.875 is the conditional complement, Pr(Y≤300 given Y<400), rather than the probability that Y exceeds 300.
Original practice · fully worked
Original variant: conditional absolute sensor deviation
A robot arm's horizontal position X, in centimeters, has distribution function F(x)=(x/10)³ for 0<x<10. A monitor records the absolute deviation D=|X-4| from a target mark. Given that D is less than 3 centimeters, calculate the probability that D exceeds 1 centimeter.
A 0.244
B 0.287
C 0.342
D 0.667
E 0.713
Variant answer in brief
The condition D<3 restricts X to (1,7), while D>1 retains the two pieces (1,3) and (5,7). Their total probability is 0.244 within a conditioning probability of 0.342, giving 122/171=0.71345 and choice E.
Setup
Setup
Translate the observed absolute-deviation bound into an interval for the original position.
D<3⟺∣X−4∣<3
D<3⟺1<X<7
Model
Model
Within the conditioning interval, exceeding one centimeter of deviation produces two disjoint position ranges.
D>1⟺X<3orX>5
{D>1}∩{D<3}={1<X<3}∪{5<X<7}
Compute
Compute
Use CDF differences for the two favorable pieces and divide by the probability of the full conditioning interval.
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