This Exam P sample reference tests Conditional Probability. The tail is S(x)=x⁻³. Conditioning on at least 1.5 gives 1-S(2)/S(1.5)=37/64=0.578125, choice A.
How to solve this Conditional Probability question
Setup
Setup
Integrating the Pareto-type density above a threshold gives a survival function equal to the inverse cube of that threshold.
S(x)=∫x∞3t−4dt=x−3
Model
Model
Within the conditioning event above 1.5, the probability of reaching 2 is the ratio of the two survival probabilities. The requested event below 2 is its complement.
Pr(X<2∣X≥1.5)=1−S(1.5)S(2)
Compute
Compute
The conditional survival probability beyond two is 0.421875, so the complementary probability is 0.578125.
1−1.5−32−3=1−(43)3=6437
Answer
Answer
The conditional probability is approximately 0.578, selecting choice A.
0.578(A)
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