A starting lineup in basketball consists of 2 guards, 2 forwards, and 1 center.
(a) a certain college team has on its roster 5 guards, 4 forwards, 3 centers. how many
different starting lineups can be created?
(b) suppose that mary is listed on the roster as a forward, and alice is listed on the
roaster as a center. assume that, for a given starting lineup, all players are randomly
chosen for each role among qualified players on the roster (guards are chose among
guards only, etc.). what is the probability that mary and alice will get on the
lineup?

Respuesta :

Using the combination formula, it is found that:

a) 180 different starting lineups can be created.

b) The probability that mary and alice will get on the lineup is [tex]\frac{1}{6}[/tex].

The order in which the players are chosen is not important, hence the combination formula is used to solve this question.

What is the combination formula?

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Item a:

The lineup is composed by:

  • 2 guards from a set of 5.
  • 2 forwards from a set of 4.
  • 1 center from a set of 3.

Hence the number of lineups is given by:

[tex]N = C_{5,2}C_{4,2}C_{3,1} = \frac{5!}{2!3!} \times \frac{4!}{2!2!} \times \frac{3!}{1!2!}  = 180[/tex]

Item b:

The number of lineups with Mary and Alice are given as follows:

  • 2 guards from a set of 5.
  • Mary and 1 other forward from a set of 3.
  • Alice as the center.

Hence:

[tex]n = C_{5,2}C_{3,1} = \frac{5!}{2!3!} \times \frac{3!}{1!2!} = 30[/tex]

Hence the probability is given by:

[tex]p = \frac{30}{180} = \frac{1}{6}[/tex]

The probability that mary and alice will get on the lineup is [tex]\frac{1}{6}[/tex].

More can be learned about the combination formula at https://brainly.com/question/25821700

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