11. There are eleven seniors and five juniors who are sprinters on the high school track team. The coach must select four sprinters to run the 800-meter relay race.
a. How many 4-sprinter relay teams can be formed from the group of 16 sprinters?
b. In how many ways can two seniors be chosen to be part of the relay team?
c. In how many ways can two juniors be chosen to be part of the relay team?
d. In how many ways can two seniors and two juniors be chosen to be part of the relay team?
e. What is the probability that two seniors and two juniors will be chosen for the relay team?

Respuesta :

Answer:

a) 1820

b) 55

c) 10

d) 550

e) 0.3022

Step-by-step explanation:

Data provided in the question:

Number of seniors = 11

Number of juniors = 5

n  = number of choices

r = number of person to be selected

number of ways = [tex]^nC_r[/tex]

a) 4-sprinter relay teams can be formed from the group of 16 sprinters

n = 16

r = 4

= ¹⁶C₄

= [tex]\frac{16!}{4!(16-4)!}[/tex]

= [tex]\frac{16\times15\times14\times13\times12!}{4\times3\times2\times1(12!)}[/tex]

=

b)  two seniors be chosen to be part of the relay team

n = 11

r = 2

= ¹¹C₂

= [tex]\frac{11!}{2!(11-2)!}[/tex]

= [tex]\frac{11\times10\times9!}{2\times1(9!)}[/tex]

=

c)  two juniors be chosen to be part of the relay team

n = 5

r = 2

= ⁵C₂

= [tex]\frac{5!}{2!(5-2)!}[/tex]

= [tex]\frac{5\times4\times3!}{2\times1(3!)}[/tex]

= 10

d)  two seniors and two juniors be chosen to be part of the relay team

=  two seniors be chosen to be part of the relay team ×  two juniors be chosen to be part of the relay team

= 55 × 10

= 550

e) P( two seniors and two juniors will be chosen for the relay team )

= [  two seniors and two juniors be chosen to be part of the relay team ] ÷ [4-sprinter relay teams can be formed from the group of 16 sprinters ]

= 550 ÷ 1820

= 0.30219 ≈ 0.3022

By finding the different numbers of combinations using C(N, K), we will get:

  • a) C =  1,820
  • b) C  = 55
  • c) C =  3
  • d) C = 165
  • e) P = 0.091

How to find the number of combinations?

If we have a set of N elements, the number of different sets of K elements that we can make out of these N, is:

[tex]C(N, K) = \frac{N!}{(N - K)!*K!}[/tex]

a) We want to select 4 runners out of a group of 16, so we have:

[tex]C(16, 4) = \frac{16!}{(16 - 4)!*4!} = \frac{16*15*14*13}{4*3*2} = 1,820[/tex]

b) We want to select 2 elements out of a group of 11, so we have:

[tex]C(11, 2) = \frac{11!}{(11 - 2)!*2!} = \frac{11*10}{2} = 55[/tex]

c) Now we want to select 2 elements out of a group of 5:

[tex]C(5, 2) = \frac{5!}{(5 - 2)!*2!} = 3[/tex]

d) Here we need to take the product of the two previous numbers of combinations:

C = 55*3 = 165

e) The probability will be given by the quotient between the combinations that have 2 seniors and 2 juniors (the number we got above) and the total number of combinations. This is:

P = 165/1,820 = 0.091

If you want to learn more about probability, you can read:

https://brainly.com/question/251701