Independent solution

How to solve this Conditional Probability question

Setup

Setup

Normalize the beta-type density first. Integrating over the unit interval gives normalizing constant five.

1=k∫01(1−y)4dy,k=51=k\int_0^1(1-y)^4dy,\qquad k=5

Model

Model

Integrating the density above a threshold gives the fifth power of one minus that threshold. Because the event above 0.4 is contained in the event above 0.1, the conditional probability is the ratio of those two survival values.

S(y)=∫y15(1−t)4dt=(1−y)5S(y)=\int_y^15(1-t)^4dt=(1-y)^5

Compute

Compute

The numerator and denominator are the fifth powers of 0.6 and 0.9. Their ratio is approximately 0.131687.

Pr⁡(Y>0.4∣Y>0.1)=0.650.95=0.131687\Pr(Y>0.4\mid Y>0.1)=\frac{0.6^5}{0.9^5}=0.131687

Answer

Answer

The conditional probability is approximately 0.1317, which rounds to 0.13 and selects choice B.

0.13(B)\boxed{0.13\quad\text{(B)}}