Now suppose the roster has 5 guards, 5 forwards, 3 centers, and 2 "swing players" (x and y) who can play either guard or forward. if 5 of the 15 players are randomly selected, what is the probability that they constitute a legitimate starting lineup

Question

Mathematics

Melvin Young

7 February, 19:18

Now suppose the roster has 5 guards, 5 forwards, 3 centers, and 2 "swing players" (x and y) who can play either guard or forward. if 5 of the 15 players are randomly selected, what is the probability that they constitute a legitimate starting lineup

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Cumulus · Answer 1

Given that a starting lineup in basketball consists of two guards, two forwards, and a center. Suppose the roster has 5 guards, 5 forwards, 3 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 15 players are randomly selected, what is the probability that they constitute a legitimate starting lineup?

The possible selections are as follows:

No swing player is selected: 5C2 x 5C2 x 3C1 = 10 x 10 x 3 = 300

One swing is selected as a guard: 5C1 x 2C1 x 5C2 x 3C1 = 5 x 2 x 10 x 3 = 300

One swing is selected as a forward: 5C1 x 2C1 x 5C2 x 3C1 = 5 x 2 x 10 x 3 = 300

One swing is selected as a guard and the other as a forward: 5C1 x 2C1 x 1C1 x 5C1 x 3C1 = 5 x 2 x 5 x 3 = 150

Two swings is selected as guards: 5C0 x 2C2 x 5C2 x 3C1 = 1 x 1 x 10 x 3 = 30

Two swings is selected as forwards: 5C0 x 2C2 x 5C2 x 3C1 = 1 x 1 x 10 x 3 = 30

Number of legitimate ways = 300 + 300 + 300 + 150 + 30 + 30 = 1110

Total number of ways = 15C5 = 3003

Therefore, probability of legitimate team = 1110 / 3003 = 0.37 or 37%.