Joint Distribution with Conditional Probability

Question

I am able to understand all the math after the 2nd step of the problem, I just don't understand where they got the 4 from in the second step of the problem. I know they are using Bayes formula, but am not sure where the 4 comes from in step 2.

Answer 1

We have $$P(Y \leq \frac{X}{4}) = \int_0^1 \int_0^{\frac{x}{4}} 10x^2 y \, \, dy \, \, dx $$ So then $$P(Y \leq \frac{X}{2}) = \int_0^1 \int_0^{\frac{x}{2}} 10x^2 y \, \, dy \, \, dx = \int_0^1 \frac{5}{4}x^4 \,\, dx = \frac{1}{4}$$ And for the step you are confused about: $$\frac{P(Y \leq \frac{X}{4}, Y \leq \frac{X}{2})}{P(Y \leq \frac{X}{2})} = \frac{P(Y \leq \frac{X}{4})}{\frac{1}{4}} = 4 P ( Y \leq \frac{X}{4})$$ with $P(Y \leq \frac{X}{4}, Y \leq \frac{X}{2}) = P(Y \leq \frac{X}{4})$ as $Y$ being less than $\frac{X}{4}$ implies adding on the constraint of also being less than $\frac{X}{2}$ does not change the probability for $0 \leq X \leq 1$.