Ron--Please help

[dddanny2006]

If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?

24/91
45/91
2/3
67/91
84/91

This is how I solve it

Possible combinations =15*14*13*12*11*10*9*8*7*6*5*4 =455
12!

To get atleast 2/3 males in the panel there are 3possibilities

1)8M and 4F
2)9M and 3F
3)10M and 2F

For the first case let the 8 males be already chosen.We only need to find the number of ways in which the females can be chosen. Since we got to select 4 females out of 5 we have 5*4*3*2=5
4!

For the second case let the 9 males be already chosen.We only need to find the number of ways in which the 3 females can be chosen. Since we got to select 3 females out of 5 we have
5*4*3 =10
3!

For the third case let the 10 males be already chosen.We only need to find the number of ways in which the females can be chosen. Since we got to select 2 females out of 5 we have
5*4 =10
2!

Thus 10+10+5=25

Favorable combinations=25/455

I know Im wrong,but why??Look at this problem below where my approach has been used.

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates
for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4
consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

a.25
b.35
c.150
d.210
e.300


First, recognize that you need to solve for project manager (PM), team leader (TL), and consultants (C) separately:

PM: one slot, two people = 2

TL: one slot, three people = 3

C: four slots, seven people = 7 * 6 * 5 * 4. With the slot method, you need to then stop and realize that there are 4! ways to rearrange the four people you have chosen. So you need to divide 840 by 24 to get 35 unique teams that can be created. Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities. If these two people are selected, we have two slots remaining with five other people to fill those slots, so there are 5 * 4 = 20, different ways to select the two other people, divided by 2, because there are 2! ways to choose those two other people. This means there are 35 - 10 = 25 consultant teams that can be selected.

Overall: 2 * 3 * 25 = 150

Now if we look at the bold'd text, we select the two trouble makers directly and then finding the possible combos for the other 2.

Going by the way the jury problem is solved,this should have been solved like this

2(Trouble makers) and 2(Non-controversial)


7*6 =21
2



and 5* 4
2

21*10=210

In the jury problem we choose 8males from 10 available males,9males from 10 available males and so on.......

Why cant we choose the 2 troublemakers out of the 7 people we have?The two could be any among the 7.


Thanks

Dan

[RonPurewal]

The difference is that the people who refuse to work together are specific people. So, for those 2 specific people, there are not different combinations of people.
Regardless of whether the language of the problem is painstakingly written to spell out this idea, it's basically just common sense. (I.e., it can't be true that any arbitrarily chosen pair of people will refuse to work together. If that were true, then no one on the committee could form any kind of team at all.)

On the other hand, this:

For the first case let the 8 males be already chosen.
...
For the second case let the 9 males be already chosen.
...
For the third case let the 10 males be already chosen.

^^ You can't do these things -- because these people are NOT already chosen. They can be any male candidates in the selection pool.

[dddanny2006]

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates
for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4
consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

a.25
b.35
c.150
d.210
e.300

Lets directly move to

C: four slots, seven people = 7 * 6 * 5 * 4. With the slot method, you need to then stop and realize that there are 4! ways to rearrange the four people you have chosen. So you need to divide 840 by 24 to get 35 unique teams that can be created. Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities.


Ill use the slot method here,order doesnt matter at all since all are going in to the same committee.

2*1*5*4 = 5/3
4*3*2*1


35-(5/3) should be our answer for the consultants,rather it is 35-10=25.

Why is this wrong?

[RonPurewal]

Hi,
We've been allowing posts from you in violation of the forum rules for some time now, but this is the point at which that stops.

1/
Please read the forum rules. First post in every folder.

2/
Please re-post your question, in conformance with those rules:
* One thread per problem. If you are asking about two problems, you should post in two different threads.
* Please title the thread as described in the forum rules. ("Ron please help" is not an appropriate title, and will make the discussion harder for future searchers to find.)

Thanks.