Question

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

Suppose the roster has 3 guards, 5 forwards, 3 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 13 players are randomly selected, what is the probability that they constitute a legitimate starting lineup? (Round your answer to three decimal places.)

Answer 1

total possible lineups =(13C5)=1287

legitimate lineups:
N(without x and y)=			(3C2)*(5C2)*(3C1)=90
x as guard			(3C1)*(5C2)*(3C1)=90
Y as guard			(3C1)*(5C2)*(3C1)=90
xas forward			(3C2)*(5C1)*(3C1)=45
both X and Y as guard			(3C0)*(5C2)*(3C1)=30
both X and Y as forward			(3C2)*(5C0)*(3C1)=9
Yas forward			(3C2)*(5C1)*(3C1)=45
one guard and other forward			(3C1)*(5C1)*(3C1)=45
total number of ways=444
P(legitimate lineups) =444/1287=0.345