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probably question - please explain!!

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probably question - please explain!!

by gmatpup » Tue Nov 08, 2011 6:25 pm
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and boys?

A. 3/8

B. 1/4

C. 3/16

D. 1/8

E. 1/16


Can someone please give me a step-by-step explanation as probability is my weakest area? (any probability tips would be very appreciated!)

Thanks so much!! :)
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by gmattaker2408 » Tue Nov 08, 2011 6:42 pm
I'm not sure. If i have to guess. I think 1/4 considering .5 probably of each girl or boy. 1/2*1/2*1/2*1/2=4/16 or 1/4

what is the correct answer?
Thanks!
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by gmatpup » Tue Nov 08, 2011 6:44 pm
Oops, forgot to post the answer, sorry!


Answer is: A
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by user123321 » Tue Nov 08, 2011 7:10 pm
if they have 4 children, the total possibilities are
2*2*2*2 = 16 possibilities
since we need to have two children out of four to be boy, we have 4c2 = 6 possibilities
so probability = 6/16 = 3/8

user123321
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Want to do it right the first time.
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by gmatpup » Wed Nov 09, 2011 10:47 am
Could you pleasee elaborate on this part:

since we need to have two children out of four to be boy, we have 4c2 = 6 possibilities
so probability = 6/16 = 3/8


How did you come up with the 6 possibilities? Thanks again!!
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by user123321 » Wed Nov 09, 2011 11:26 am
gmatpup wrote:Could you pleasee elaborate on this part:

since we need to have two children out of four to be boy, we have 4c2 = 6 possibilities
so probability = 6/16 = 3/8


How did you come up with the 6 possibilities? Thanks again!!
out of four children, exactly two should be boys. Here order is not important. If there are two boys among the four children, then that should be fine. so it is like selecting two places among four available spaces and placing boys there.hence 4C2 = 6
(we are not required to worry about placing girls because automatically girls will come in remaining two places in 2c2 = 1 way).

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by gmatpup » Wed Nov 09, 2011 11:35 am
I know this sounds silly so I apologize for asking another question but I did the 4C2 (or at least I thought I did) and I did not get 6...

Would you mind doing a step by step for me? :(
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by user123321 » Wed Nov 09, 2011 1:04 pm
Sure no problem.

lets use slot method. We have four blanks. each blank has two ways of filling either M(ale) or F(emale).
M/F M/F M/F M/F
so total possibilities are 2*2*2*2 = 16 possibilities.

The problem says out of those four slots, two slots should be M and other two slots should be F.
choose two slots out of four available slots and place 2M there = 4C2 = 4!/(2!.2!) = 24/4 = 6
assign remaining two slots with 2F = 2C2 = 1
so total possibilities are 6*1 = 6
so probability is 6/16 = 3/8

For your understanding I will give what those 6 possibilities are.
MMFF
MFFM
FFMM
MFMF
FMMF
FMFM

this is nothing but arranging four objects of which two are alike = 4!/2!.2! = 6

user123321
HTH.

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