Independent solution

How to solve this Uniform Distribution question

Setup

Setup

Because the lower endpoint is below three, the first conditional tail ratio depends only on the upper endpoint.

Pr⁡(X>4∣X>3)=b−4b−3=0.90\Pr(X>4\mid X>3)=\frac{b-4}{b-3}=0.90

Model

Model

Solve the first length ratio, then express the second conditional probability using its truncated interval.

b−4=0.90(b−3)b-4=0.90(b-3)
Pr⁡(X>4∣X<11)=11−411−a=0.70\Pr(X>4\mid X<11)=\frac{11-4}{11-a}=0.70

Compute

Compute

The equations give both endpoints; use the full support length for the unconditional tail.

b=13,711−a=0.70⇒a=1b=13,\qquad \frac{7}{11-a}=0.70\Rightarrow a=1
Pr⁡(X>4)=13−413−1=912=0.75\Pr(X>4)=\frac{13-4}{13-1}=\frac{9}{12}=0.75

Answer

Answer

The unconditional probability is 0.75.

0.75(B)\boxed{0.75\quad\text{(B)}}