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Chapter 2: Q42E (page 74)

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

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.)

b. Now suppose the roster has \({\rm{5}}\) guards, \({\rm{5}}\) forwards, \({\rm{3}}\)centers, and \({\rm{2}}\) “swing players” (X and Y) who can play either guard or forward. If 5 of the \({\rm{15}}\) players are randomly selected, what is the probability that they constitute a legitimate starting lineup?

Short Answer

Expert verified

a) The number of ways to select different starting line ups is \({\rm{72}}{\rm{.}}\)

b) The probability is \({\rm{0}}{\rm{.3696}}\)

Step by step solution

01

Introduction

The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

02

Calculating lineup can be created

A lineup of two guards, two forwards, and a center has a total of five players.

(a):

A collegiate squad consists of three centers, four guards, four forwards, and one guy (X) who can play forward or guard. We'll investigate how many possible starting lineups can be formed in three distinct scenarios using the clue from the Exercises.

1. Lineup without\({\rm{X}}\):

We want to use:

Product Rule for k-Tuples

Assume that we can select first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways, and that for each selected element we can select the second element in \({{\rm{n}}_{\rm{2}}}\)ways, then we have the number of pairs to be\({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\).

Similarly, for ordered collection of \({\rm{k}}\)elements, where the \({{\rm{k}}^{{\rm{th }}}}\)can be chosen in \({{\rm{n}}_{\rm{k}}}\)ways, then there are \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\)possible \({\rm{k}}\)-tuples.

03

Calculation for lineup can be created

Here, \({{\rm{n}}_{\rm{i}}}\)will be the number of ways to select players for particular position. For example,

\({{\rm{n}}_{\rm{1}}}{\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right){\rm{ = 3}}\)

is the number of ways we can select one center out of the three in the team. We use combinations here because the order is not important.

A combination is an unordered subset. For \({\rm{n}}\)individuals in a group, the number of combinations size \({\rm{k}}\)is denoted as \({{\rm{C}}_{{\rm{k,n}}}}\)or \(\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right)\)which we read as from \({\rm{n}}\)elements we choose\({\rm{k}}\). The proposition:

\(\begin{aligned}{{\rm{C}}_{{\rm{k,n}}}} \rm &= \left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\\ \rm &= \frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}\\ \rm &= \frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\end{aligned}\)

04

Calculation for lineup can be created

We also have that

\(\begin{aligned}{{\rm{n}}_{\rm{2}}} \rm &= \left( {\begin{aligned}{\rm{4}}\\{\rm{2}}\end{aligned}} \right)\\ \rm &= \frac{{{\rm{4!}}}}{{{\rm{2!(4 - 2)!}}}}\\ \rm &= 6 \end{aligned}\)

where we choose \({\rm{2}}\) out of four guards, and

\(\begin{aligned}{{\rm{n}}_{\rm{3}}} \rm &= \left( {\begin{aligned}{\rm{4}}\\{\rm{2}}\end{aligned}} \right)\\ \rm &= \frac{{{\rm{4!}}}}{{{\rm{2!(4 - 2)!}}}}\\ \rm &= 6 \end{aligned}\)

is the number of ways to select \({\rm{2}}\) forwards out of four the team has.

The number of ways to select different starting line ups is

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{*}}{{\rm{n}}_{\rm{3}}}{\rm{ = 3*6*6 = 108}}\)

05

Calculation for lineup can be created

\({\rm{2}}\). Lineup with \({\rm{X}}\)as guard:

We do the identical calculations as before; however, we only have one less guard to select from because one is fixed. It solely affects \({{\rm{n}}_{\rm{2}}}\), which is now \({{\rm{n}}_{\rm{3}}}\).

\(\begin{array}{c}{{\rm{n}}_{\rm{2}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{4}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{4!}}}}{{{\rm{1!(4 - 1)!}}}}\\ \rm &=& 4 \end{array}\)

because we have \({\rm{4}}\) guards left and we need only one more.

The number of ways to select different starting line ups is

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{*}}{{\rm{n}}_{\rm{3}}}{\rm{ = 3*4*6 = 72}}{\rm{.}}\)

06

Calculating probability

(b):

To calculate the probability that the five constitute a legitimate starting lineup, we want to use

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A }}}}{{{\rm{\# of outcomes in the sample space }}}}\)

where we need to find # of favorable outcomes in \({\rm{A}}\) and # of outcomes in the sample space.

We will calculate the number of favorable outcomes the same way we calculated in (a), with taking in consideration that there are two "swing players". Reminder, in this team we have \({\rm{\;5}}\) guards, \({\rm{\;5}}\) forwards, \({\rm{3}}\) centers, and \({\rm{2}}\) "swing players".

07

Calculation

With the same logic as in (a), we will have a sum of

- number of ways to create starting line up without \({\rm{X}}\) or \({\rm{Y}}\);

- number of ways to create starting line up with \({\rm{X}}\) as guard and without \({\rm{Y}}\);

- number of ways to create starting line up with \({\rm{X}}\) as forward and without \({\rm{Y}}\);

- number of ways to create starting line up with \({\rm{Y}}\) as guard and without \({\rm{X}}\);

- number of ways to create starting line up with \({\rm{Y}}\) as forward and without \({\rm{X}}\);

- number of ways to create starting line up with \({\rm{X}}\) as guard and \({\rm{Y}}\) as guard;

- number of ways to create starting line up with \({\rm{X}}\) as guard and \({\rm{Y}}\) as forward;

- number of ways to create starting line up with \({\rm{X}}\) as forward and \({\rm{Y}}\) as guard;

- number of ways to create starting line up with \({\rm{X}}\) as forward and \({\rm{Y}}\) as forward.

08

Calculation

\(\begin{array}{c}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}*\frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}*\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 10*10*3 &=& 300;* \\{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}*\frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}*\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 5*10*3 &=& 150;* \end{array}\)

\(\begin{array}{c}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}*\frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}*\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 10*5*3 &=& 15{{\rm{0}}^{\rm{*}}}\\{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}*\frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}*\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 5*10*3 &=& 15{{\rm{0}}^{\rm{*}}}^{\rm{*}}\end{array}\)

\(\begin{array}{c}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}*\frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}*\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 10*5*3 &=& 150{{\rm{;}}^{\rm{*}}}\\{\rm{1*}}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& 1*\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right)*\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& 1* \frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}{\rm{*}}\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 1*10*3 &=& 30 {{\rm{;}}^{\rm{*}}}\end{array}\)

\(\begin{array}{c}{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right){\rm{*}}\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right){\rm{*}}\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}{\rm{*}}\frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}{\rm{*}}\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 5*5*3 &=& 75;*\\{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,5}}}}{\rm{*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right){\rm{*}}\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right){\rm{*}}\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}{\rm{*}}\frac{{{\rm{5!}}}}{{{\rm{1!(5 - 1)!}}}}{\rm{*}}\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\\ \rm &=& 5*5*3 &=& 75;\end{array}\)

\(\begin{array}{c}{{\rm{C}}_{{\rm{2,5}}}}{\rm{*1*}}{{\rm{C}}_{{\rm{1,3}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right){\rm{*1*}}\left( {\begin{array}{*{20}{l}}{\rm{3}}\\{\rm{1}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{5!}}}}{{{\rm{2!(5 - 2)!}}}}{\rm{*1*}}\frac{{{\rm{3!}}}}{{{\rm{1!(3 - 1)!}}}}\end{array}\)

Therefore, the number of favorable outcomes is

\({\rm{300 + 4*150 + 2*30 + 2*75 = 1110}}\)

09

Calculation

When order is not crucial, the number of outcomes in the sample space equals the number of ways to take players out of \({\rm{15}}\) (combinations again)

The probability that the five constitute a legitimate starting lineup is

\(\begin{array}{c} \rm P(A) &=& \frac{{{\rm{\# of favorable outcomes in A }}}}{{{\rm{\# of outcomes in the sample space }}}}\\ \rm &=& \frac{{{\rm{1110}}}}{{{\rm{3003}}}}\\ \rm &=& 0{\rm{.3696}}\end{array}\)

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Most popular questions from this chapter

A quality control inspector is examining newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let p denote the probability that the flaw is detected during any one fixation (this model is discussed in “Human Performance in Sampling Inspection,” Human Factors, \({\rm{1979: 99--105)}}{\rm{.}}\)

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b. Give an expression for the probability that a flaw will be detected by the end of the nth fixation.

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