Probability: Choosing players from a group.

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July 22nd 2010, 03:48 AM
phillychum

Probability: Choosing players from a group.

some homework question.

A total of 6 players is selected at random to form a volleyball team from a group of 10 girls and 5 boys, 6 of the girls and 3 of the boys have played the game before. Find the probability that of the 6 selected
a) 4 are girls and 2 are boys
b) at least 4 of the players are girls that have played the game before.
July 22nd 2010, 04:28 AM
mathaddict

Quote:

Originally Posted by phillychum

some homework question.

A total of 6 players is selected at random to form a volleyball team from a group of 10 girls and 5 boys, 6 of the girls and 3 of the boys have played the game before. Find the probability that of the 6 selected
a) 4 are girls and 2 are boys
b) at least 4 of the players are girls that have played the game before.

For (a) Calculate (10C4 x 5C2)/15C6

(b) sorry, i misinterpreted this as among the 6 selected players who are experienced, at least 4 of the players are girls that have played the game before. Thanks Soroban.
July 22nd 2010, 05:57 AM
Soroban

Hello, phillychum!

Quote:

There are 10 girls and 5 boys who play volleyball.
A team of 6 players are selected at random
6 of the girls and 3 of the boys have played the game before.

Some players are Experienced, others are New to the game.

. . $\begin{array}{c||c|c||c|} & \text{Exp} & \text{New} & \text{Total} \\ \hline \hline \text{Girls} & 6 & 4 & 10 \\ \hline \text{Boys} & 3 & 2 & 5 \\ \hline \hline \text{Total} & 9 & 6 & 15 \\ \hline \end{array}$

Quote:

Find the probability that of the 6 selected

a) 4 are girls and 2 are boys

There are: . ${15\choose6}$ possible teams.

4 girls, 2 boys: . ${10\choose4}{5\choose2}$ teams.

Therefore: . $P(\text{4G 2B}) \:=\:\dfrac{{10\choose4}{5\choose2}}{{15\choose6}} \:=\:\dfrac{2100}{5005} \:=\:\dfrac{60}{143}$

Quote:

b) at least 4 of the players are girls that have played the game before.

There are 6 Experienced girls $(G^*)$ and 9 Others.

There are three cases to consider:

. . $\begin{array}{cccccc}(1) & 4\;G^*,\:2\text{ Others:} & {6\choose4}{9\choose2} &=& 540 \\ \\[-3mm] (2) & 5\:G^*,\:1\text{ Other:} & {6\choose5}{9\choose1} &=& 54 \\ \\[-3mm] (3) & 6\;G^*,\:0\text{ Other:} & {6\choose6}{9\choose0} &=& 1 \\ \\[-4mm] \cline{3-5} & & \text{Total:} & & 595 \end{array}$

Therefore: . $P(\text{at least 4 }G^*) \;=\;\dfrac{595}{5005} \;=\;\dfrac{17}{143}$