Respuesta :
Answer:
The probability that the committee will consist of 1 from each type of monster is 0.1318 and Suppose that the Pokemon refuse to be on the same committee as a the Trolls.So,the probability that this type of committee is formed is 0.5357
Step-by-step explanation:
No. of Trolls = 3
No. of Nazguls =4
No. of Ents = 4
No. of Pokemon = 5
Total Monsters = 16
We are given that A committee of 4 is to be picked to represent the group at the Monster Bash.
(a) Find the probability that the committee will consist of 1 from each type of monster.
So, the probability that the committee will consist of 1 from each type of monster:
= [tex]\frac{^3C_1 \times ^4C_1 \times ^4C_1 \times ^5C_1}{^{16}C_4}[/tex]
= [tex]\frac{\frac{3!}{1!(3-1)!} \times \frac{4!}{1!(4-1)!} \times \frac{4!}{1!(4-1)!}\times \frac{5!}{1!(5-1)!}}{\frac{16!}{4!(16-4)!}}[/tex]
= [tex]\frac{\frac{3!}{1!(2)!} \times \frac{4!}{1!(3)!} \times \frac{4!}{1!(3)!}\times \frac{5!}{1!(4)!}}{\frac{16!}{4!(12)!}}[/tex]
= [tex]0.1318[/tex]
b)Suppose that the Pokemon refuse to be on the same committee as a the Trolls.
So, the probability that this type of committee is formed
= [tex]\frac{(^3C_3 \times ^8C_1)+(^3C_2 \times ^8C_2)+(^3C_1 \times ^8C_3)+(^3C_0 \times ^8C_4)+(^5C_4 \times ^8C_0)+(^5C_3 \times ^8C_1)+(^5C_2 \times ^8C_2)+(^5C_1 \times ^8C_3)}{^{16}C_4}[/tex]
= [tex]\frac{(\frac{3!}{3!(3-3)!} \times \frac{8!}{1!(8-1)!})+(\frac{3!}{2!(3-2)!} \times\frac{8!}{2!(8-2)!})+(\frac{3!}{1!(3-1)!} \times\frac{8!}{3!(8-3)!})+(\frac{3!}{0!(3-0)!} \times \frac{8!}{4!(8-4)!})+(\frac{5!}{4!(5-4)!} \times \frac{8!}{0!(8-0)!})+(\frac{5!}{3!(5-3)!} \times \frac{8!}{1!(8-1)!})+(\frac{5!}{2!(5-2)!} \times\frac{8!}{2!(8-2)!})+(\frac{5!}{1!(5-1)!} \times \frac{8!}{3!(8-3)!})}{\frac{16!}{4!(16-4)!}}[/tex]
= [tex]0.5357[/tex]
Hence The probability that the committee will consist of 1 from each type of monster is 0.1318 and Suppose that the Pokemon refuse to be on the same committee as a the Trolls.So,the probability that this type of committee is formed is 0.5357
