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, three forwards, and one individual (X) who can play either guard or forward. How many different starting lineups can be created? lineups (b) Now suppose the roster has 3 guards, 5 forwards, 4 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 14 players are randomly selected, what is the probability that they constitute a legitimate starting lineup?

Explanation / Answer

Answer to part a)

The team has 3 centers, 5 guards, 3 forwards and one X who can be either a guard or forward

we need : 2 guards, 2 forwards & 1 center

.

Number of ways in which it can be formed is:

Case I: if X is not chosen

N(I) = 3C1 * 5C2 * 3C2 = 3*10*3 = 90

.

Case II: if X becomes a guard

N(II) = 3C1 * 6C2 * 3C2 = 3*15*3 = 135

.

Case III: if X becomes a formward

N(III) = 3C1 *5C2 *4C2 = 3 * 10 * 6 = 180

.

Thus total number of combinations = N(I) + N(II) + N(III)

Total number of combinations = 90+135+180 = 405

.

Answer to Part b)

3 guards, 5 forwards , 4 centers and 2 swing players

.

either the two players become : Guards

then we gt 5 guards, 5 forwards , 4 centers

P(team) =5C2 * 5C2 *4C1 / 14C5

P(team) = 10 * 10 *4 / 2002 = 0.1998

.

either th two palyers become: forwards

then we get 3 guards, 7 forwards , 4 centers

P(team) = 3C2 * 7C2 * 4C1 / 14C5 = 3 * 21 *4 / 2002 = 0.1259

.

or we get one as Guard and one as forward

then we get: 4 guards, 6 forwards and 4 centers

P(team) =2 * 4C2 * 6C2 * 4C1 / 14C5

[we mutiply this one by 2, because either X can be guard and Y can be forward, or Y can be guard and X can be forward]

P(team) = 2 * 6 * 15 * 4 / 2002 = 0.3596

.

Total Probability = 0.1998 + 0.1259 + 0.3596 = 0.6853

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