A starting lineup in basketball consists of two guards, two forwards, and a cent
ID: 3205407 • Letter: A
Question
A starting lineup in basketball consists of two guards, two forwards, and a center. A certain college team has on its roster three centers, five guards, four forwards, and one individual (X) who can play either guard or forward. How many different starting lineups can be created? lineups Now suppose the roster has 5 guards, 5 forwards, 3 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 15 players are randomly selected, what is the probability that they constitute a legitimate starting lineup? ()Explanation / Answer
A.a) Given,
line up should consist of,
Guards
Forward
Center
2
2
1
members in the team
Guards
Forward
Center
5
4
3
and one individual(X) who can play either guard or forward
total possible combinations?
total line-ups=line up without X+ line-up with X as guard+ line-up with X as forward
total line-ups =5C2*4C2*3C1+1*5C1*4C2*3C1+1*5C2*4C1*3C1
total line-ups =180+90+120=390
therefore, total possible line-ups are 390
A.b) Given,
line up should consist of,
Guards
Forward
Center
2
2
1
members in the team
Guards
Forward
Center
5
5
3
Now, we have 2 swing players (X and Y) who can either play guard or forward
using X and Y we can have 9 different combinations. they are,
X
0
0
0
G
G
G
F
F
F
Y
0
G
F
0
G
F
0
G
F
sum of legal combinations = (5C2*5C2*3C1)+(5C1*5C2*3C1)+(5C2*5C1*3C1)+(5C1*5C2*3C1)+(5C0*5C2*3C1)+(5C1*4C1*3C1)+( 5C2*5C1*3C1)+( 5C1*4C1*3C1)+(5C2*5C0*3C1)
sum of legal combinations =300+150+150+150+30+60+150+60+30
sum of legal combinations =1080
total combinations=15C5=3033
probability of having legitimate team= (sum of legal combinations)/ (total combinations)
probability of having legitimate team =1080/3033
probability of having legitimate team =0.3596
Guards
Forward
Center
2
2
1
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