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

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

Also Read :   What happens if you put too much water in jello mix?

(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,

Also Read :   Which sample when dissolved in 1.0 liter of water, produces a solution with the lowest freezing point?

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.

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.

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