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) A certain college team has on its roster three 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.] Incorrect: Your answer is incorrect. lineups

(b) Now suppose the roster has 3 guards, 4 forwards, 3 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 12 players are randomly selected, what is the probability that they constitute a legitimate starting lineup? (Round your answer to three decimal places.) Incorrect: Your answer is incorrect.

Explanation / Answer

Solution:-   So we have:
3 C,
4 G,
4 F,
1 F / G

a) Follow the hint:
....... C .... F ....... G
No x: 3 * 4c2 * 4c2 = 3 * 6 * 6 = 108
x = F: 3 * 4c1 * 4c2 = 3 * 4 * 6 = 72
X = G: 3 * 4c2 * 4c1 = 72

Total = (108 + 72 +72) = 252

b) Total possible lineups = 12c5 = 12!/ (5!(125)!) =   792

Pr(ok lineup) = 252/792 = 28/88 = 7 / 22 = 0.31818

NOTE:::     Based on part (b) it appears that the problem description is missing "four forwards", to bring the total to 12.

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