Question

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

(a) A certain college team has on its roster four centers, four guards, three 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, 5 forwards, 4 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? (Round your answer to three decimal places.)

Answer 1

college team has,

centers = 4

guards = 4

forwards = 3

and X = either guard or forward

a) the no. of different lineups = no. of lineups without X + no. of lineups with X as guard + no.of lineups with X as forward

= 4C2*3C2*4C1 + 4C1*3C2*4C1 + 4C2*3C1*4C1 = 72 + 48 + 72 = 192

b) 4 guards, 5 forwards, 4 centers

number of lineups without X:

number of lineups without y, with y as a guard, with y as a forward = 4C2*5C2*4C1 + 4C1*5C2*4C1 + 4C2*5C1*4C1 = 360 + 160 + 120 = 640

number of lineups with X as guard:

number of lineups without y, with y as a guard, with y as a forward = 4C1*5C2*4C1 + 4C0*5C2*4C1 + 4C1*5C1*4C1 = 160 + 40 + 80 = 280

number of lineups with X as forward:

number of lineups without y, with y as a guard, with y as a forward = 4C2*5C1*4C1 + 4C1*5C1*4C1 + 4C2*5C0*4C1 = 120 + 80 + 24 = 224

total number of ways to select 5 out of 15 = 15C5 = 3003

probability that they form a legitimate starting lineup = (640 + 280 + 224)/3003 = 0.381