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Chapter 2: Q33E (page 73)

Again, consider a Little League team that has \({\rm{15}}\) players on its roster.

a. How many ways are there to select \({\rm{9}}\) players for the starting lineup?

b. How many ways are there to select \({\rm{9}}\)players for the starting lineup and a batting order for the \({\rm{9}}\) starters?

c. Suppose \({\rm{5}}\) of the \({\rm{15}}\) players are left-handed. How many ways are there to select \({\rm{3}}\) left-handed outfielders and have all \({\rm{6}}\) other positions occupied by right-handed players?

Short Answer

Expert verified

a) \({\rm{1,816,214,400}}\)

b) \({\rm{659,067,881,472,000;}}\)

c) \({\rm{9,072,000}}\)

Step by step solution

01

team roaster consists of

A team's roster consists of 15 players.

02

determining ways to select 9 players for the starting lineup 

We utilize permutations since the order is crucial (we have distinct positions) and we have \({\rm{15}}\) players and \({\rm{9}}\) positions for players.

An ordered subset is referred to as a permutation. The number of permutations of size \({\rm{k}}\) for \({\rm{n}}\)persons in a group is represented as \({{\rm{P}}_{{\rm{k,n}}}}\).The proposal is this:

\({{\rm{P}}_{{\rm{k,n}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{(n - k)!}}}}\)

We have some possibilities too,

\({{\rm{P}}_{{\rm{9,15}}}}{\rm{ = }}\frac{{{\rm{15!}}}}{{{\rm{(15 - 9)!}}}}{\rm{ = }}\frac{{{\rm{15!}}}}{{{\rm{6!}}}}{\rm{ = 1,816,214,400}}\)

03

 Determining the ways are there to select 9 players for the starting lineup and a batting order for the 9 starters

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\)ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

The ordered pair in this case is a starting lineup in first place and a batting order in second place. We can pick a starting lineup in \({\rm{1,816,214,400}}\)ways (a), and we can choose the batting order in \({\rm{1,816,214,400}}\)ways

\({{\rm{P}}_{{\rm{9,9}}}}{\rm{ = }}\frac{{{\rm{9!}}}}{{{\rm{(9 - 9)!}}}}{\rm{ = }}\frac{{{\rm{9!}}}}{{{\rm{1!}}}}{\rm{ = 362,880}}\)

We need permutations of size \({\rm{n = 9}}\)for \({\rm{k = 9}}\)people. The ultimate solution, based on everything we've discussed, is

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{ = 1,816,214,400*362,880 = 659,067,881,472,000}}{\rm{.}}\)

04

Determining the ways to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players

The sequence is still important, so there are a few options.

\({{\rm{P}}_{{\rm{3,5}}}}{\rm{ = }}\frac{{{\rm{5!}}}}{{{\rm{(5 - 3)!}}}}{\rm{ = 60}}\)

three left-handed athletes to pick from and

\({{\rm{P}}_{{\rm{6,10}}}}{\rm{ = }}\frac{{{\rm{10!}}}}{{{\rm{(10 - 6)!}}}}{\rm{ = 151,200}}\)

strategies to choose six out of ten left-handed athletes for the remaining six spots From:

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\)ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{ = 60*151,200 = 9,072,000}}{\rm{.}}\)

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Most popular questions from this chapter

An engineering construction firm is currently working on power plants at three different sites. Let Aidenote the event that the plant at site i is completed by the contract date. Use the operations of union, intersection, and complementation to describe each of the following events in terms of \({A_1}\), \({A_2}\), and \({A_3}\), draw a Venn diagram, and shade the region corresponding to each one.

a. At least one plant is completed by the contract date.

b. All plants are completed by the contract date.

c. Only the plant at site 1 is completed by the contract date.

d. Exactly one plant is completed by the contract date.

e. Either the plant at site 1 or both of the other two plants are completed by the contract date.

A box in a supply room contains \({\rm{15}}\) compact fluorescent lightbulbs, of which \({\rm{5}}\) are rated \({\rm{13}}\)-watt, \({\rm{6}}\)are rated \({\rm{18}}\)-watt, and \({\rm{4}}\) are rated \({\rm{23}}\)-watt. Suppose that three of these bulbs are randomly selected.

a. What is the probability that exactly two of the selected bulbs are rated \({\rm{23}}\)-watt?

b. What is the probability that all three of the bulbs have the same rating?

c. What is the probability that one bulb of each type is selected?

d. If bulbs are selected one by one until a \({\rm{23}}\)-watt bulb is obtained, what is the probability that it is necessary to examine at least 6 bulbs?

The accompanying table gives information on the type of coffee selected by someone purchasing a single cup at a particular airport kiosk.

Small

Medium

Large

Regular

\(14\% \)

\(20\% \)

\(26\% \)

Decaf

\(20\% \)

\(10\% \)

\(10\% \)

Consider randomly selecting such a coffee purchaser.

a. What is the probability that the individual purchased a small cup? A cup of decaf coffee?

b. If we learn that the selected individual purchased a small cup, what now is the probability that he/she chose decaf coffee, and how would you interpret this probability?

c. If we learn that the selected individual purchased decaf, what now is the probability that small size was selected, and how does this compare to the corresponding unconditional probability of (a)?

Show that for any three events \({\rm{A,B}}\), and \({\rm{C}}\) with \({\rm{P(C) > 0}}\),\({\rm{P(A}} \cup {\rm{B}}\mid {\rm{C) = P(A}}\mid {\rm{C) + P(B}}\mid {\rm{C) - P(A}} \cap {\rm{B}}\mid {\rm{C)}}\).

Suppose that 55% of all adults regularly consume coffee,45% regularly consume carbonated soda, and 70% regularly consume at least one of these two products.

a. What is the probability that a randomly selected adult regularly consumes both coffee and soda?

b. What is the probability that a randomly selected adult doesn鈥檛 regularly consume at least one of these two products?
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