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?
To find the probability that a legitimate starting lineup can be formed, we need to determine the total number of possible outcomes (sample space) and the number of favorable outcomes.
First, let's calculate the total number of possible outcomes. We are selecting 5 players out of 14, so we can use the combination formula:
Total outcomes = C(14, 5) = 14! / (5! * (14 - 5)!) = 2002
Next, let's determine the number of favorable outcomes, i.e., the number of ways to form a legitimate starting lineup.
Since there are specific requirements for each position, we need to consider different cases:
1. Selecting 2 guards, 2 forwards, and 1 center:
Number of ways to select 2 guards = C(5, 2) = 5! / (2! * (5 - 2)!) = 10
Number of ways to select 2 forwards = C(4, 2) = 4! / (2! * (4 - 2)!) = 6
Number of ways to select 1 center = C(3, 1) = 3! / (1! * (3 - 1)!) = 3
2. Selecting 1 guard, 2 forwards, 1 center, and 1 swing player:
Number of ways to select 1 guard = C(5, 1) = 5! / (1! * (5 - 1)!) = 5
Number of ways to select 2 forwards = C(4, 2) = 4! / (2! * (4 - 2)!) = 6
Number of ways to select 1 center = C(3, 1) = 3! / (1! * (3 - 1)!) = 3
Number of ways to select 1 swing player = C(2, 1) = 2! / (1! * (2 - 1)!) = 2
3. Selecting 2 guards, 1 forward, 1 center, and 1 swing player:
Number of ways to select 2 guards = C(5, 2) = 5! / (2! * (5 - 2)!) = 10
Number of ways to select 1 forward = C(4, 1) = 4! / (1! * (4 - 1)!) = 4
Number of ways to select 1 center = C(3, 1) = 3! / (1! * (3 - 1)!) = 3
Number of ways to select 1 swing player = C(2, 1) = 2! / (1! * (2 - 1)!) = 2
Now, we sum up the favorable outcomes for each case:
Total favorable outcomes = (10 * 6 * 3) + (5 * 6 * 3 * 2) + (10 * 4 * 3 * 2) = 360 + 180 + 240 = 780
Finally, we can calculate the probability:
Probability = (Total favorable outcomes) / (Total outcomes) = 780 / 2002 ≈ 0.389 (rounded to three decimal places)
Therefore, the probability that a legitimate starting lineup is formed when randomly selecting 5 players is approximately 0.389.
528/2002 = 0.264 but it's wrong.
Both answers of 0.4735 and 0.3686 are wrong. We got 768 cases, 0.3836, still wrong.
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
Is this the correct answer?
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?
180+150+150+30+18+90+120 = 738/2002 = .3686
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.