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.)

Question

Mathematics

Larry Beard

9 January, 14:12

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.)

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Emmalee Sellers · Answer 1

You are given a roster with 3 guards, 5 forwards, 3 centers, and 2 "swing players" (x and y) who can play either guard or forward. You are given a condition that 5 of the 13 players are randomly selected. You are asked to find the probability that they constitute a legitimate starting lineup.

For these cases, there are a lot so,

legitimate ways:

two swings and one gurad = 2C2*5C2*3C1 = 30

two swings used as forwards = 3C2*2C2*3C1 = 9

two swings used one guard = 2C1*3C1*1C1*5C1*3C1 = 90

one swing used as forward = 3C2*2C1*5C1*3C1 = 90

zero swing used = 3C2*5C2*3C1 = 90

total of legitimate ways = 489

Total ways = 13C5 = 1287

The probability that they constitute a legitimate lineup is = 489/128 = 0.38