Assume that there are 14 board members: 9 females, and 5 males including Mark. There are 3 tasks to be assigned. Note that assigning the same people different tasks constitutes a different assignment.1) Find the probability that both males and females are given a task.2) Find the probability that Mark and at least one female are given tasks.

Answer:Total members = 14 Total tasks = 3So, total outcomes = [tex]^{14}C_3[/tex]1) Find the probability that both males and females are given a task.No. of females = 9No. of males = 5Favorable events = 2 female 1 male + 2 male 1 male = [tex]9C_2 \times 5C_1 +9C_1 \times 5C_2[/tex]So,  the probability that both males and females are given a task: = [tex]\frac{^9C_2 \times ^5C_1 +9C_1 \times ^5C_2}{^{14}C_3}[/tex]= [tex]\frac{\frac{9!}{2!(9-2)!} \times \frac{5!}{1!(5-1)!} +\frac{9!}{1!(9-1)!} \times \frac{5!}{2!(5-2)!}}{\frac{14!}{3!(14-3)!}}[/tex]= [tex]0.7417[/tex]So,  the probability that both males and females are given a task is 0.74172)Find the probability that Mark and at least one female are given tasks.Since mark is fixed , so places are left So, favorable events =  2 female + 1 female 1 male = [tex]^9C_2 +^4C_1 \times ^9C_1[/tex]So,  the probability that Mark and at least one female are given tasks. :=  [tex]\frac{^9C_2 +^4C_1 \times ^9C_1}{^{13}C_2}[/tex]=  [tex]\frac{\frac{9!}{2!(9-2)!} +\frac{4!}{1!(4-1)!} \times \frac{9!}{1!(9-1)!} }{\frac{13!}{2!(13-2)!}} [/tex]=  [tex]0.9230[/tex]So, The probability that Mark and at least one female are given tasks is 0.9230