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

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? 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

Given:- 5-G,5-f,3-Center,2-Sp

p(constitute a legimate starting lineup)=total no.of lineup with X and Y/15C5

lineup without X and Y,=3C1*5C2*5C2=3*10*10=300

lineup when X sit and Y play =3C1*5C1*5C1+3C1*5C2*5C1=3*5*10+3*10*5=300

lineup when Y sit and X play =3C1*5C1*5C2+3C1*5C2*5C1=3*5*10+3*10*5=300

lineup when X and Y both play=3C1*5C0*5C2+3C1*5C2*5C0+3C1*5C1*5C1=3*1*10+3*10*1+3*5*5=135

total no.of lineup with X and Y =300+300+300+135=1035

15C5=3003

therefore,required probability=1035/3003=0.345

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