1:2 points ! My Notes O Ask Your A starting lineup in basketball consists of two

Question

1:2 points ! My Notes O Ask Your A starting lineup in basketball consists of two guards, two forwards, and a center (a) A certain college team has on its roster four centers, five guards, five forwards, and one individual (X) who can play either guard or forward. How many dferent starting ineups can be created? [Hint: Consider lineups without x, then lineups with X as quard, then lineups with X as forward.) lineugs (b] Now suppose the roster has 4 guards, forwards, 4 centers, and 2 swing players" (X and y) who can play alther guard or forward. If 5 of the 15 players are randomly selected, what is the probabilty that they constitute a legitimate stating lineup? (Round your answer to three decimal places.) 679 Need Help?RasdTat to Tutere My Notes Ask Your 9 - points

Explanation / Answer

Answer to the question is as follows:

a.

4 C, 5G, 5F, 1X - G/F

Basketball line up as 2G, 2F, 1C

a.

total types of line up =

without X - 5C2*5C2*4C1 = 400

with X as G - 5C1*5C2*4C1 = 200

with X has F - 5C1*5C2*4C1 = 200

total = 800

b. P( 5 out of 15 players form legit line ups)

case 1 both are not playing:

4C2*5C2*4C1 = 6*10*4 = 240

case 2 X as gaurd and Y as forward

4C1*5C1*4C1 = 4*5*4 = 100

case 3 X has forward and Y as gaurd

4C1*5C1*4C1 = 4*5*4 = 100

case 4 X has gaurd and Y not playing

4C1*5C2*5C1 = 200

case 5 X not playing and Y as gaurd

4C1*5C1*5C2 = 200

case 6 X as forward and Y not playing

4C2*5C1*4C1 = 6*5*4 = 120

case 7 X not playing and Y as forward

5C1*4C2*4C1 = 120

total combos = 15C5

Hence, the probability is (240+100*2 + 120*2 +200*2) /15C5  

= .3596

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