A starting lineup in basketball consists of two guards, two

forwards, and a center. (a) A certain college team has on its roster 3 centers, 4

guards, 5 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 4 guards, 4 forwards, 4 centers, and

2 “swing players” (X and Y) who can play either guard or forward.

If 5 of the 14 players are randomly selected, what is the

probability that they constitute a legitimate starting lineup?

Hint:

General guidance

Concepts and reason

Combination: A combination is a selection of objects without considering the order.

Probability: The ratio of the number of favorable outcomes to certain event and total number of possible outcomes is called as the probability of an event.

Fundamentals

Let n is the total number of observation and r is the number of success then the combination is defined as,

The probability of an event is defined as,

Step-by-step

Step 1 of 2

(a)

The number of different starting lineups is obtained below:

From the given information, the roster has 3 centers, 4 guards, 5 forwards and one individual (X) who can play either guard or forward. Moreover, lineup starts with two guards, two forwards and a center.

If X is not in lineup, then the possible number ways for lineup is .

If X play as forward, then remaining one forward is selected from 5. Therefore, the possible line ups are .

If X play as guard, then remaining one guard is selected from 4 guards. Therefore, the possible line ups are .

Thus, the possible different lineups are .

Part a

The number of different starting lineups is 390.

The number of different starting lineups is obtained by adding the number of lineups without X, the number of lineups with X as forward and the number of lineups with X as guards.

Use to find the probability that the randomly selected players constitute a legitimate starting lineup.

Step 2 of 2

(b)

Form the given information, 5 players are selected from 14 players randomly. The total number for selecting 5 players is . Moreover, there are 4 guards, 4 forwards, 4 centers and 2 “swing players” (X and Y) who can play either guard or forward.

The number of lineups without X and Y is .

If X is selected and play as guard, then the possible lineups are .

If X is selected and play as forward, then the possible lineups are .

If Y is selected and play as guard, then the possible lineups are .

If Y is selected and play as forward, then the possible lineups are .

If X and Y are both selected, here X plays as guard and Y plays as forward then possible lineups are .

If X and Y are both selected, here X plays as forward and Y plays as guard then possible lineups are .

If X and Y are selected and both plays as guards. Then possible lineups are .

If X and Y are selected and both plays as forward. Then possible lineups are .

Hence, the possible number of legitimate lineups is,

Therefore, the required probability is,

Part b

The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.

The probability that the randomly selected players constitute a legitimate starting lineup is obtained by dividing the number of possible legitimate lineups by the total number of ways for selecting 5 players.

Answer

Part a

The number of different starting lineups is 390.

Part b

The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.

Answer only

Part a

The number of different starting lineups is 390.

Part b

The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.

“C, (n-r)!r!

Number of favorable outcomes for an event Probability = Total number of outcomes

“C, (n-r)!r!

(1)-(*)(2)-180

01-0H(3)-20

(1)-(*)-(1)-120

180+90+120=390

“C, (n-r)!r!

zooʻz=(6V)

(9)_02_(2)=144

(1)C-()-26

(1)C-()-26

(1)C-()-26

(1)C-()-26

(1)_07-01-04

(1)_07-01-04

(1)_02_0=24

(1)_02_0=24

14+6+6+6+6+4+4+24+24 = 704

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