Independent solution

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−3S(x)=\int_x^\infty3t^{-4}dt=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(2)S(1.5)\Pr(X<2\mid X\ge1.5)=1-\frac{S(2)}{S(1.5)}

Compute

Compute

The conditional survival probability beyond two is 0.421875, so the complementary probability is 0.578125.

1−2−31.5−3=1−(34)3=37641-\frac{2^{-3}}{1.5^{-3}}=1-\left(\frac34\right)^3=\frac{37}{64}

Answer

Answer

The conditional probability is approximately 0.578, selecting choice A.

0.578(A)\boxed{0.578\quad\text{(A)}}