Asked by jil

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

Now suppose the roster has 5 guards, 4 forwards, 3 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?
Answered by Reiny
number of ways to choose any 5 of the 14
= C(14,5) = 2002

number of "legitimate" starting lineups

Cases:
GGFCF
1. X and Y excluded:
C(5,2) x C(3,1) x C(4,2) = 180

GXFCF
2. X plays guard, Y included in the forwards
C(5,1) x C(5,2) x C(3,1) = 150

GYFCF
3. Y plays guard, X included in the forwards
same as #2 ---> 150

GGXCY
4. Both X and Y play forward
C(5,2) x 1 x C(3,1) = 30

XYFCF
5. Both X and Y play guard
1 x C(4,2) x C(3,1) = 18

I will let you finish it.
Answered by jil
528/2002 = 0.264 but it's wrong.
Answered by Reiny
ok jil, I forgot the case where only one of "swing players" is in the lineup

6. X plays guard, Y does not play
GXFCF
C(5,1) x 1 x C(4,2) x C(3,1) = 90

7. X plays forward, Y does not play
GGXCF
C(5,2) x 1 x C(4,1) x C(3,1) = 120

We have the same cases if Y plays, but not x

so the total of all exceptions
= 180 + 2(150) + 30 + 18 + 2(90+120) = 948


prob = 948/2002 = .4735

How does that look?
Answered by jil
180+150+150+30+18+90+120 = 738/2002 = .3686
Answered by joseph
Both answers of 0.4735 and 0.3686 are wrong. We got 768 cases, 0.3836, still wrong.
Answered by Anonymous
My answer is different

Legitimate lineup when both x and y are guards comes out to be 3c1*7c2*5c2=630

Similar would be the case when both are forwards

Also when one is forward and one is guard ,no. Of ways are (3c1*6c2*6c2)*2 since their positions are interchangeable

Total ways= 2610

Ways of selecting 5 out of 15 players = 15c5 = 3003

Required probability is 2610/3003=
0.869

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