Independent solution

How to solve this Continuous Distributions question

Setup

Setup

The normalizing constant cancels from the conditional ratio, so integrate the linear density only over the relevant intervals.

Pr⁡(T≤10∣T>5)=Pr⁡(5<T≤10)Pr⁡(T>5)\Pr(T\le10\mid T>5)=\frac{\Pr(5<T\le10)}{\Pr(T>5)}

Model

Model

An antiderivative of kt is kt squared over two. Apply it to the target interval and the full conditioned interval.

Pr⁡(T≤10∣T>5)=(k/2)(102−52)(k/2)(202−52)\Pr(T\le10\mid T>5)=\frac{(k/2)(10^2-5^2)}{(k/2)(20^2-5^2)}

Compute

Compute

Cancel the common factor and simplify the squared-endpoint differences.

102−52202−52=75375=15=0.20\frac{10^2-5^2}{20^2-5^2}=\frac{75}{375}=\frac15=0.20

Answer

Answer

The conditional probability is 0.20.

0.20(B)\boxed{0.20\quad\text{(B)}}