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Probability Question

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Probability Question

by jfranco23 » Mon Jan 19, 2009 12:02 pm
I found this question, can anyone give me the answer and the explanation?

- From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
A. 101
B. 94
C. 21
D. 53
E. 32
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by jfranco23 » Mon Jan 19, 2009 12:59 pm
Sorry, that are not the answer choices, the right ones are:

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
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Re: Probability Question

by aroon7 » Mon Jan 19, 2009 8:00 pm
jfranco23 wrote:I found this question, can anyone give me the answer and the explanation?

- From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
A. 101
B. 94
C. 21
D. 53
E. 32
we need 2 Boys AND 2 girls
so it is
(3/6)(2/5)(3/4)(2/3) = 1/10

ie A

let me explain in more detail:

assume that the sequence is BBGG (as we finally multiply everything, need not worry about the order)

probability of choosing first boy = 3/6 (3 boys out of total 6)
probability of choosing second boy = 2/5 (one boy is already selected so, 2 boys out of total 5)
similarly,
probability of choosing first girl = 3/4
probability of choosing second girl = 2/3

we have an AND condition. so multiply the above numbers to get 1/10

HTH
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by goblue2010 » Tue Jan 20, 2009 3:44 pm
yes, but i believe you need to multiply 1/10 by (4!/2!2!) = 6, since there are several ways in which you can have 2 boys and 2 girls.

So I believe the correct answer is actually 6 * 1/10 = 6/10 = 3/5
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by robbie » Thu Jan 22, 2009 12:42 pm
goblue2010, I think you're right also. I understand why you multiplied by six, because they're are six possibilities for two boys and two girls, but how did you come up with (4!/2!2!)? I understand that it is six, but I don't see where you pulled the factorials from.

Thanks in advance for the explanation.
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by goblue2010 » Fri Jan 23, 2009 11:45 am
I'm not totally sure, but I think this is the explanation.

You have 4 things that can be picked out. If order matters, there are 4! ways to do it.

However, if you have any sub-categories within those 4 things, then it would be 4! divided by the # of things in each sub-category. Here it's 2 boys and 2 girls, so that's why it's 4!/2!2!

If you get asked, how many ways can you order A, A, B, C? i think the right answer is 4!/2!, for the same reason. You have 2 A's, but only 1 B and 1 C.

if someone could verify my explanation, that would be great.

thanks
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by robbie » Fri Jan 23, 2009 10:17 pm
There was a response here by gaggleofgirls that seems to have vanished.... if you've been following this then you'll understand my post.


Reading through and knowing of Carrie's other posts, I think she is on to something.


As for the new spin on the question:


(5/7)(4/6)(2/5)(1/4)= 1/21


Carrie, let me know how I did!



Thanks everyone for all the effort.
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by aroon7 » Fri Jan 23, 2009 10:25 pm
What is the OA?
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What is the OA ?

by chintudave » Fri Jan 23, 2009 10:44 pm
IMO - D 3/5

I do not see this problem as conditional probability.

Total number of combination - 6C4 i.e 15

Total number of combination for two boys - 3C2 i.e 3
Total number of combination for two girls - 3C2 i.e 3

So total number of 2 boys and 2 girls is 3C2 * 3C2 = 9

So probability is 9/15 = 3/5 = D
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by gaggleofgirls » Fri Jan 23, 2009 11:19 pm
Robbie - I deleted the posts because I realize that I was wrong or at least had left something out.

I like the way chintudave responded.

There are 6! / 4!*2! possibilities for a subgroup of 4 out of a group of 6 = 15

For probability, we need #desired outcomes / # possible outcomes.
So, 15 is the # possible outcomes (denominator).

Then the number of desired outcomes is when we have 2 boys and 2 girls, no matter what order they were chosen in.

So, the probability of getting 2 of the 3 boys is 3! / 2! *1! = 3
same for the probability of getting 2 of the 3 girls = 3
Probability of 2 girls AND 2 boys (and is the tipoff for multiplication) is 3*3 = 9

So 9/15 = 3/5

So, for my modified problem of choosing the 4 person equal gender subset from 5 boys and 2 girls....

total outcomes is 7! / 4! *3! = 35. This is the denominator.

Now, to get 2 boys from the 5 boys is 5! / 2! * 3! = 10

To get 2 girls from the 2 girls, there is only 1 possibility.

10 * 1 = 10. This is the numerator.

10/35 = 2/7

The way I was doing it (that got me the 1/10 for the original problem or 1/21 for this modified problem) was actually assigning order when there is no order.

The probability in this modified problem that you will get
B B G G is 1/21 (5/7 * 4/6 * 2/5 * 1/4)

But that is only one instance of an equal gender subgroup of 4.

B G B G is another 1/21

Now, we have an OR
B B G G or B G B G or any of the other ways you can chose 2 boys and 2 girls, so you need to add up all the 1/21st tha get you equal gender.

Turns out that there are 6 of them

B B G G
B G B G
B G G B
G G B B
G B G B
G B B G

1/21 + 1/21 + 1/21 + 1/21 + 1/21 + 1/21 = 6/21 = 2/7

For the original problem, you have these same 6 desired outcomes that each have 1/10 probability so:
1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 = 6/10 = 3/5

What I didn't like in the answer was the 4!/2!*2! to get 6.

-Carrie
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by robbie » Fri Jan 23, 2009 11:31 pm
You definitely put in the time. Thanks again, I think that I finally got it.
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by gaggleofgirls » Fri Jan 23, 2009 11:39 pm
Time is my problem now - with enough time, I can get to the right answer.

Now to get it down to 2 minutes :D

-Carrie
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by blitz28 » Fri Jan 23, 2009 11:51 pm
There is only 1 way of selecting equal number of boys and girls

For boys- 3C2
For Girls- 3C2

Thus probability from a total possible selection 6C4

= (3C2*3C2)/6C4 = 3/5

OA- 3/5
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by aroon7 » Sat Jan 24, 2009 10:51 am
blitz28 wrote:There is only 1 way of selecting equal number of boys and girls

For boys- 3C2
For Girls- 3C2

Thus probability from a total possible selection 6C4

= (3C2*3C2)/6C4 = 3/5

OA- 3/5
Thanks Blitz
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by peddisetty » Sun Feb 15, 2009 1:07 pm
Why cant we pick 3 boys 3 girls and 1 boy 1 girl ?
Why is it only two 2 boys and 2 girsl?.

In the question it is asked, equal no of boys and equal no of girls.

Please somebody clarify.

Thanks.
Raj Peddisetty
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