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, five 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. lineups (b) Now 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.)

Explanation / Answer

3 centers

4 guards

5 forwards

1 X(guard or forward)

(a)

Lineup without X, 3C1*4C2*5C2 = 180

if X is selected as guard, possible lineups are 3C1*4C1*5C2 = 120

if X is selected as forward, possible lineups are 3C1*4C2*5C1 = 90

Total distinct possible lineups are 390

b)

Possible lineups without X and Y are 3C1*3C2*5C2 = 90

if X and Y both are gurads, possible lineups are 3C1*5C2 = 30

if X and Y both are forwards, possible lineups are 3C1*3C2 = 9

if X is guard and Y is forward, possible lineups are 3C1*3C1*5C1 = 45

if Y is guard and X is forward, possible lineups are 3C1*3C1*5C1 = 45

total possible lineups are 90 + 30 + 9 + 45 + 45 = 219

Total random lineups are 13C5 = 1287

Hence required probabiltiy = 219/1287 = 0.17016

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.