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a) In how many ways can this be done? b) The coach received last minute funding to allow 3 additional players.Is how ways can he select the 3 additional players? c) One of the players is Tim. The probability that he will not be among the 8 selected?

a group of 5 basketball players is to be selected from a basketball team of 13 players to travel and promote the team.

See below:

Explanation:

a

We can choose 5 players from a population of 13 by using a combination (order of choosing isn't important):

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

$\left(\begin{matrix}13 \\ 5\end{matrix}\right) = 1287$

b

We can now pick 3 additional players from the $13 - 5 = 8$ remaining:

$\left(\begin{matrix}8 \\ 3\end{matrix}\right) = 56$

c

There are $\left(\begin{matrix}13 \\ 5\end{matrix}\right) = 1287$ ways the players not going on the trip can be chosen. Of those, if we force Tim to be one of the players not going, the remaining way to chose those four remaining spots out of 12 remaining players is:

$\left(\begin{matrix}12 \\ 4\end{matrix}\right) = 495$

This means that the probability of his being a part of the group not going on the trip is:

$\frac{495}{1287} \approx 0.3846$