# Probability Problems

• December 16th 2010, 01:49 AM
johnsy123
Probability Problems
1)Three men and three women are to be randomly seated in a row. Find the probability that both the end places will be filled by women.

2)A netball team of seven players is to be chosen from six men and seven women. Find the probability that the selected team contains more men than women.
• December 16th 2010, 02:00 AM
Prove It
1. The probability of the first end seat being a woman is $\displaystyle \frac{3}{6}$ because there are 3 women.

Note that once a woman has been seated there, there are only 2 women of the 5 remaining people left. So the probability of the second end seat being a woman is $\displaystyle \frac{2}{5}$.

Therefore, the probability of both end seats being filled by women is $\displaystyle \frac{3}{6} \times \frac{2}{5} = \frac{1}{5}$.

2. You need the probability of having 4 men... Each of the terms below represents each selection...

$\displaystyle \frac{6}{13} \times \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{3}{143}$.
• December 16th 2010, 06:35 AM
Soroban
Hello, johnsy123!

Quote:

$\text{Three men and three women are to be randomly seated in a row.}$
$\text{Find the probability that both the end places will be filled by women.}$

There are: . $6! \,=\,720$ possible seatings.

There are women on each end: . $W\;\square\,\square\,\square\,\square\;W$

There are 3 choices for the woman on the left end.
There are 2 choices for the woman on the right end.
The remaining 4 people can be seated in $4!$ ways.

There are: . $3\cdot2\cdot4! \:=\:144$ ways.

Therefore: . $P(\text{woman on both ends}) \:=\:\dfrac{144}{720} \:=\:\dfrac{1}{5}$

Quote:

$\text{2) A team of 7 players is to be chosen from 6 men and 7 women.}$
$\text{Find the probability that the team contains more men than women.}$

There are: . $_{13}C_7 \:=\: {13\choose7} \:=\:1716$ possible teams.

$\begin{array}{ccccccc}
\text{4 men, 3 women:} & {6\choose4}{7\choose3} &=& 525 \\ \\[-3mm]
\text{5 men, 2 women:} & {6\choose5}{7\choose2} &=& 126 \\ \\[-3mm]
\text{6 men, 1 woman:} & {6\choose6}{7\choose1} &=& \;\;\;7 \\ \\[-3mm]\hline \\[-4mm]
& \text{Total:} && 658 \end{array}$

There are 658 teams with more men than women.

Therefore: . $P(\text{more men}) \;=\;\dfrac{658}{1716} \;=\;\dfrac{329}{858}$