Let Y1 and Y2 denote the proportion of time during which employees I and II actually performed their assigned tasks during a workday, The joint density of Y1 and Y2 is given by
f(y1,y2) = { y1+y2, 0<=y1<=1, 0<=y2<=1, 0, elsewhere

Required:
a. Find the marginal density function of Y1 and Y2
b. Find P(Y1 >= 1/2 | Y2 >= 1/2).
c. If employee II spends exactly 50% of the dayworking on assigned duties, find the probability that employee I spends more than 75% of the day working on similarduties.

[tex]f_{(y_1,y_2)} \left \{ {{y_1+y_2, \ \ 0\ \le \ y_1 \ \le 1 , \ \ 0 \ \ \le y_2 \ \ \le 1} \atop {0, \ \ \ elsewhere \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \right.[/tex]

a)To find the marginal density of [tex]y_1[/tex].

[tex]f_{y_1} (y_1) = \int \limits ^{\infty}_{-\infty} f_{y_1,y_2} (y_1 >y_2) \ dy_2[/tex]

[tex]=\int \limits ^{1}_{0}(y_1+y_2)\ dy_2[/tex]

[tex]=\int \limits ^{1}_{0} \ \ y_1dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2[/tex]

[tex]= y_1 \ \int \limits ^{1}_{0} dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2[/tex]

[tex]= y_1[y_2]^1_0 + \bigg [ \dfrac{y_2^2}{2}\bigg]^1_0[/tex]

[tex]= y_1 [1] + [\dfrac{1}{2}][/tex]

[tex]= y_1 + \dfrac{1}{2}[/tex]

i.e.

[tex]f_{(y_1}(y_1)}= \left \{ {{y_1+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.[/tex]

The marginal density of [tex]y_2[/tex] is:

[tex]f_{y_1} (y_2) = \int \limits ^{\infty}_{-\infty} fy_1y_1(y_1-y_2) dy_1[/tex]

[tex]= \int \limits ^1_0 \ y_1 dy_1 + y_2 \int \limits ^1_0 dy_1[/tex]

[tex]=\bigg[ \dfrac{y_1^2}{2} \bigg]^1_0 + y_2 [y_1]^1_0[/tex]

[tex]= [ \dfrac{1}{2}] + y_2 [1][/tex]

[tex]= y_2 + \dfrac{1}{2}[/tex]

i.e.

[tex]f_{(y_1}(y_2)}= \left \{ {{y_2+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.[/tex]

b)

[tex]P\bigg[y_1 \ge \dfrac{1}{2}\bigg |y_2 \ge \dfrac{1}{2} \bigg] = \dfrac{P\bigg [y_1 \ge \dfrac{1}{2} . y_2 \ge\dfrac{1}{2} \bigg]}{P\bigg[ y_2 \ge \dfrac{1}{2}\bigg]}[/tex]

[tex]= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} f_{y_1,y_1(y_1-y_2) dy_1dy_2}}{\int \limits ^1_{\frac{1}{2}} fy_1 (y_2) \ dy_2}[/tex]

[tex]= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} (y_1+y_2) \ dy_1 dy_2}{\int \limits ^1_{\frac{1}{2}} (y_2 + \dfrac{1}{2}) \ dy_2}[/tex]

[tex]= \dfrac{\dfrac{3}{8}}{\dfrac{5}{8}}[/tex]

[tex]= \dfrac{3}{8}}\times {\dfrac{8}{5}}[/tex]

[tex]= \dfrac{3}{5}}[/tex]

= 0.6

(c) The required probability is:

[tex]P(y_2 \ge 0.75 \ y_1 = 0.50) = \dfrac{P(y_2 \ge 0.75 . y_1 =0.50)}{P(y_1 = 0.50)}[/tex]

[tex]= \dfrac{\int \limits ^1_{0.75} (y_2 +0.50) \ dy_2}{(0.50 + \dfrac{1}{2})}[/tex]

[tex]= \dfrac{0.34375}{1}[/tex]

= 0.34375