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Probability Q - using factorials

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Probability Q - using factorials

by justbeatit » Thu Aug 07, 2008 10:42 pm
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?

Answ: 67/91

How do u use the factorial formula here?

Many thanks in advance.
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by AKS-AKS » Fri Aug 08, 2008 6:04 am
Initially

There are 2/3 men means 2/3 *15 = 10 men
There are 1/3 women means 1/3 * 15 = 5 women

now we need to choose 12 jury. we can do it in 15C12 way

In jury team (that needs to be selected) we need to 2/3 men=
2/3 * 12 = 8 men

So women would be 4. If in new team there are more than 4 women condition will not be satisfied. which means If we have 5 women and 7 men we cannot choose new jury team.

We can choose 7 men and 5 women in 10C7 *5C5 ways
= 10C7*1

Now probability that new team could not be chosen is
10C7 / 15C12 = 24/91

Probability that new jury team could be chosen is 1 - 24/91 = 67/91

So answer is 67/91
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by justbeatit » Fri Aug 08, 2008 6:14 am
thanks a lot!

just on the topic of probability...

are u able to distinguish when to use factorials ro solve or solve by 'manual thinking' ie listing the elements and going through by hand?

it seems that probability questions with higher numbers require using factorials.... but i was wondering if anyone was able to tell when to use either of the methods?

thanks!
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by Senator 153 » Fri Aug 08, 2008 1:38 pm
The question asks what is the probability of getting "at least" a two-thirds men's jury. It would seem that your answer checks the probability of getting exactly a 2/3rds male jury... not more than that. Your math checks out, I'm not secong-guessing the answer, but can you elaborate on why it checks out?
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by AKS-AKS » Sun Aug 10, 2008 7:27 am
There are 2 ways in which you can approach for the solution.

One direct other indirect

The procedure expalined above is indirect one.

If you follow direct aproach

Here comes the solution

New jury team needs to have 12 membes in which 2/3 rd should be male members.

So number of male members could be 8 or 9 or 10.
If you choose 8 members U have to choose 4 female members

If 9 male members are chosen than 3 female members are to be chosen

If 10 male members are chosen than 2 female members are to be chosen

So total favourable case is 10C8 * 5C4 + 10C9 * 5C3 +10C10 * 5C2
= 225 + 100 + 10
The Event case is 15C12 = 7*5*13

Probability that at least 8 male members are chosen out of 12 members is (favourable case )/ (Event case) = 335/455 = 67/91

I hope this helps..
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Trying to figure this out another way

by midgely » Thu Mar 05, 2009 12:29 am
I understand that you do 1 - (prob. of 7 men, 5 women), I just can't figure out why it doesn't work to calculate this probability like this:

(10 men to pick from /15 ppl total) * (8 men.../14 ppl) * (7 m/13 ppl) ...*(5 women to pick from /8 ppl remaining) * (4 women/7 ppl) * ... (1 woman remaining / 4 ppl remaining)

I have used this method for other things like "What's the probability of drawing four white socks out of a drawer with 4 identical black socks and 6 identical white socks?"

Is my arithmetic wrong (quite possible, I suck at arithmetic and often make careless errors), or can I not use this method for this type of problem? Why not?

I keep getting the following answer for the probability of 7 men, 5 women on a jury of 12 chosen from a pool of 10 men, 5 women (15 people total): 1/3*7*11*13 or 1/3003. Therefore the prob of at least 8 men is 1 - 1/3003 which is 3002/3003.

Logically, I realize this are too good of odds - it's really not a near certainty that there could be 5 women on the jury, so I know something is wrong - what am I missing? Is the only way to do this problem really to find out the # of combinations of 5 women, 7 men and divide that into the total # of ways to choose 12 people from 15??

Thanks for the help!
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