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, five 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.] 600 Correct: Your answer is correct. 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.)

Explanation / Answer

A.a) let’s assume 3 scenarios where X can be

Guard

Forward

Not playing

also, we have 3 centrs, 5 guards, 5 forwards

case 1: X will be guard

total combinations are

3C1*5C1*5C2=150

case 2: X will be forward

total combinations are

3C1*5C2*5C1=150

case 3: X not playing

3C1*5C2*5C2=300

hence total combinations are 150+150+300=600

A.b) we have 3 centrs, 3 guards, 4 forwards

2 swing players X and Y

case1: Both not playing

total combinations are

3C2*4C2*3C1=54

case2:X as guard and Y as forward

total combinations are

3C1*4C1*3C1=36

case3: X as forward and Y as guard

total combinations are

3C1*4C1*3C1=36

case4: X as guard and Y not playing

total combinations are

3C1*4C2*3C2=54

case5:X not playing and Y as guard

total combinations are

3C2*4C2*3C1=54

case6: X as forward and Y not playing

total combinations are

3C2*4C1*3C1=36

case7: X not playing and Y as forward

total combinations are

3C2*4C1*3C1=36

total legitimate combinations are 54+36+36+54+54+36+36=306

total combinations possible are 12C5=792

hence the probability is 306/792=0.386 or 38.6%

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