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Probability and Statistics for Eng...

9th Edition
Jay L. Devore
ISBN: 9781305251809

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BuyFindarrow_forward

Probability and Statistics for Eng...

9th Edition
Jay L. Devore
ISBN: 9781305251809
Textbook Problem

A starting lineup in basketball consists of two guards, two forwards, and a center.

  1. a. A certain college team has on its roster three centers, four guards, four forwards, and one individual (X) who can play either guard or forward. How many different starting lineups can be created? [Hint: Consider lineups without X, then lineups with X as guard, then lineups with X as forward.]
  2. b. Now suppose the roster has 5 guards, 5 forwards, 3 centers, and 2 “swing players” (X and Y) who can play either guard or forward. If 5 of the 15 players are randomly selected, what is the probability that they constitute a legitimate starting lineup?

a.

To determine

Find the total number of starting lineups.

Explanation

Given info:

The starting lineup in basketball which has 2 guards, 2 forwards and 1 center. A college has on its roster three centers, four guards, four forwards, and individual (X) who can play either guard or forward.

Calculation:

Combination:

A combination is a selection of objects without regard to order. It is denoted as Ck,n

The expression for the combination is,

Ck,n=n!k!(nk)!

Here, n represents the group size and k represents the subset size.

The number of teams when player X sitting out is obtained as given below:

N(player X sits out)=(31)(42)(42)=3!1!(31)!×4!2!(42)!×4!2!(42)!=3!1!×2!×4!2!(2)!×4!2!(2)!=3×6×6

=108

The number of teams when player X guards is obtained as given below

N(player X guards)=(31)(41)(42)=3!1!(31)!×4!1!

b.

To determine

Find the probability of selecting legitimate starting lineup.

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