Answer:
0.25
Step-by-step explanation:
From the given information:
Since the times are uniform distributed over the one hour period (0,1):
Then;
[tex]f_1(y_1) = 1[/tex]
[tex]f_2(y_2) = 1[/tex]
So, [tex]Y_1[/tex] and [tex]Y_2[/tex] are independent; then:
[tex]f(y_1|y_2) = f_1(y_1)f_2(y_2) \\ \\ = 1(1) = 1[/tex]
[tex]P(Y_1\le 0.5 Y_2 \le 0.5) = \int ^{0.5}_{0} \int ^{0.5}_{0} f(y_1,y_2) dy_{2}dy_{1}[/tex]
[tex]\implies \int ^{0.5}_{0} \int ^{0.5}_{0} (1) dy_{2}dy_{1}[/tex]
[tex]= \dfrac{1}{4}[/tex]
= 0.25