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?
Answer
General guidance
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.
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 
.
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.
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,
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
The number of different starting lineups is 390.
The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.
Answer only
The number of different starting lineups is 390.
The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.