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.]

I found the right answer of 600.

(b) Now suppose the roster has 4 guards, 4 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.)

I've worked out this problem about 10 times and constantly get 0.410 which is not the correct answer according to webassign.

Explanation / Answer

Here we are told to do part b)

b) Here, the number of ways that a team could be made could be computed here as:

Therefore total number of ways here is computed as: 108 + 144 + 144 + 48 + 18 + 18 = 480

Now the required probability here is computed as:

= Number of ways that team could be chosen so that it is a legit team / Total number of ways to select 5 players from 13 players

= 480 / (13c5)

= 480 / 1287

= 0.3730

Therefore 0.3730 is the required probability here.

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