Answer:
(a) f(x,y)=1 for 5≤x≤6 and 5≤y≤6 and 0 otherwise, (c) 23/144
Step-by-step explanation:
(a) Let X=Annie's arrival time and Y = Alvie's arrival time.
Since X and Y is independent with each other, we can simply write as
[tex]f(x,y)=\left \{ {{1,\ 5\leq x\leq 6,\ 5\leq y\leq 6 } \atop {0, \ otherwise}} \right.[/tex]
(b) Since X and Y independent from each other we can simply write
[tex]P(5:15\leq X \leq 5:45,5:15\leq Y \leq 5:45)= P(5:15\leq X \leq 5:45).P(5:15\leq Y \leq 5:45)[/tex]
and from the distribution of X and Y (given as uniform between 5:00 PM and 6:00 PM), we can write as
[tex]P(5:15\leq X \leq 5:45)=P(5:15\leq Y \leq 5:45)=0.5[/tex]
Therefore,
[tex]P(5:15\leq X \leq 5:45,5:15\leq Y \leq 5:45)= 0.5 * 0.5=0.25[/tex]
0.25 is the probability that they both arrive between 5:15 and 5:45.
(c) Simply the event of interest can be defined as
[tex]A=\{ {(x,y): |x-y|\leq }\frac{1}{12} \}[/tex]
This situation can be visualized as given in the attachment:
The region of interest (probability of the event) is region II.
[tex]Area\ II = 1 - Area\ I - Area \ III\\\\Area\ II = 1-\frac{1}{2}\frac{11}{12}\frac{11}{12}-\frac{1}{2}\frac{11}{12}\frac{11}{12}\\\\Area\ II = 1-\frac{121}{144} =\frac{23}{144}[/tex]
