A committee consisting of four different people, from 5 men and 5 women.

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williamrobertsuk

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Out of five men and five women, we form a committee consisting of four different people. Assuming that each committee of size four is equally likely, find the probabilities of the following events:
  1. The committee consists of two men and two women.
  2. The committee has more women than men.
  3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.
  4. Both Alice and Bob are members of the committee.
Shouldn't it be like a coin toss but with more than just heads and tails?
 
It's not like a coin toss, because each "toss" must consist of four different people, and coins don't do that. It's much more like a card game.

So, what have you tried, and where are you stuck? There are several possible methods, so it will be very helpful to see what approach(es) you consider (e.g. combinations, multiplying probabilities, ...).
 
williamrobertsuk said:
Out of five men and five women, we form a committee consisting of four different people. Assuming that each committee of size four is equally likely, find the probabilities of the following events:
  1. The committee consists of two men and two women.
  2. The committee has more women than men.
  3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.
  4. Both Alice and Bob are members of the committee.
Click to expand...
These are simple selection questions: \(\displaystyle \binom{10}{4}=\frac{10!}{(4!)([10-6]!}\), ten choose four.
That is the number of possible committees of four. Five choose four is the number of all women. so how many have at least one man?
Please post all you answers so that they may be checked.
 
Dr.Peterson said:
It's not like a coin toss, because each "toss" must consist of four different people, and coins don't do that. It's much more like a card game.

So, what have you tried, and where are you stuck? There are several possible methods, so it will be very helpful to see what approach(es) you consider (e.g. combinations, multiplying probabilities, ...).
Click to expand...
I just started with it, and I wanted some input or guidance. Thanks.
 
1. The committee consists of two men and two women.
There are 5 choices for the first man then 4 choices for the second man so 5*4= 20 choices- but for each "Frank, Bob" in that list there is also "Bob, Frank". Since the order in which they are chosen does not matter, we must divide by 2- there are 20/2= 10 choices for the two men on the committee. In exactly the same way, there are 10 choices for the two women so there are 10*10= 100 such committees.

2. The committee has more women than men.
Since there are 4 places there may be 4 women, 0 men or 3 women 1 man.
For "4 women 0 men", we can think of this as choosing the one woman who is not on the committee. There are 5 choices so 5 such committees.
For "3 women 1 man", we can think of this as choosing the two women who are not on the committee and the one man who is on the committee. As above there are 5*4/2= 10 ways to choose the two women who are not on the committee and there are 5 ways to choose the man who is on the committee. There are 10*5= 50 ways.
So there are 5+ 50= 55 committees with more women than men.

3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.
"At least one man" is the same as "not all women". Ignoring the distinction between men and women there are 10 people to choose from. There are 10 ways to choose the first person, nine the next, then 8, then 7. But there are also 4*3*2*1 orders for the same 4 people so there are 10*9*8*7/4*3*2*1= 210 4 person committees without any restriction. For an "all woman" committee there are 5 ways to chose the one woman who is not on the committee so there are 5 all woman committees. There are 210- 5= 215 committees with "at least one man".

4. Both Alice and Bob are members of the committee.
If Alice and Bob are on the committee we only have to choose the two other people on the committee from the remaining 8 people. There are 8*7/2= 28 such committees.
 
HallsofIvy said:
1. The committee consists of two men and two women.
There are 5 choices for the first man then 4 choices for the second man so 5*4= 20 choices- but for each "Frank, Bob" in that list there is also "Bob, Frank". Since the order in which they are chosen does not matter, we must divide by 2- there are 20/2= 10 choices for the two men on the committee. In exactly the same way, there are 10 choices for the two women so there are 10*10= 100 such committees.

2. The committee has more women than men.
Since there are 4 places there may be 4 women, 0 men or 3 women 1 man.
For "4 women 0 men", we can think of this as choosing the one woman who is not on the committee. There are 5 choices so 5 such committees.
For "3 women 1 man", we can think of this as choosing the two women who are not on the committee and the one man who is on the committee. As above there are 5*4/2= 10 ways to choose the two women who are not on the committee and there are 5 ways to choose the man who is on the committee. There are 10*5= 50 ways.
So there are 5+ 50= 55 committees with more women than men.

3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.
"At least one man" is the same as "not all women". Ignoring the distinction between men and women there are 10 people to choose from. There are 10 ways to choose the first person, nine the next, then 8, then 7. But there are also 4*3*2*1 orders for the same 4 people so there are 10*9*8*7/4*3*2*1= 210 4 person committees without any restriction. For an "all woman" committee there are 5 ways to chose the one woman who is not on the committee so there are 5 all woman committees. There are 210- 5= 215 committees with "at least one man".

4. Both Alice and Bob are members of the committee.
If Alice and Bob are on the committee we only have to choose the two other people on the committee from the remaining 8 people. There are 8*7/2= 28 such committees.
Click to expand...
Your results aren't correct even though your approach makes sense. :)
 
williamrobertsuk said:
Out of five men and five women, we form a committee consisting of four different people. Assuming that each committee of size four is equally likely, find the probabilities of the following events:
  1. The committee consists of two men and two women.
  2. The committee has more women than men.
  3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.
  4. Both Alice and Bob are members of the committee.
Click to expand...
1) \(\displaystyle \dfrac{\binom{5}{2}\binom{5}{2}}{\binom{10}{4}}\)

2) \(\displaystyle \dfrac{\binom{5}{3}+\binom{5}{4}}{\binom{10}{4}}\)

3) \(\displaystyle 1 -\dfrac{\binom{5}{4}}{\binom{10}{4}}\)

4) \(\displaystyle \dfrac{\binom{8}{2}}{\binom{10}{4}}\)
 
HallsofIvy said:
1. The committee consists of two men and two women.
There are 5 choices for the first man then 4 choices for the second man so 5*4= 20 choices- but for each "Frank, Bob" in that list there is also "Bob, Frank". Since the order in which they are chosen does not matter, we must divide by 2- there are 20/2= 10 choices for the two men on the committee. In exactly the same way, there are 10 choices for the two women so there are 10*10= 100 such committees.

2. The committee has more women than men.
Since there are 4 places there may be 4 women, 0 men or 3 women 1 man.
For "4 women 0 men", we can think of this as choosing the one woman who is not on the committee. There are 5 choices so 5 such committees.
For "3 women 1 man", we can think of this as choosing the two women who are not on the committee and the one man who is on the committee. As above there are 5*4/2= 10 ways to choose the two women who are not on the committee and there are 5 ways to choose the man who is on the committee. There are 10*5= 50 ways.
So there are 5+ 50= 55 committees with more women than men.

3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.
"At least one man" is the same as "not all women". Ignoring the distinction between men and women there are 10 people to choose from. There are 10 ways to choose the first person, nine the next, then 8, then 7. But there are also 4*3*2*1 orders for the same 4 people so there are 10*9*8*7/4*3*2*1= 210 4 person committees without any restriction. For an "all woman" committee there are 5 ways to chose the one woman who is not on the committee so there are 5 all woman committees. There are 210- 5= 215 committees with "at least one man".

4. Both Alice and Bob are members of the committee.
If Alice and Bob are on the committee we only have to choose the two other people on the committee from the remaining 8 people. There are 8*7/2= 28 such committees.
Click to expand...

Thank you, it is so helpful!
 
I would do #2 slightly differently. First there are 10c4 ways of choosing 4 people from 10. We know that there are 100 ways of choosing 2m and 2w as per HallsofIvy. Now exactly half of the 10c4 - 100 remaining possible committees will have more women than men
So the answer is (10c8 - 100)/2
 
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