Respuesta :
Answer:
Total members = 14
Total tasks = 3
So, total outcomes = [tex]^{14}C_3[/tex]
1) Find the probability that both males and females are given a task.
No. of females = 9
No. of males = 5
Favorable 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 0.7417
2)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
