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Probability

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by leumas » Sat Sep 10, 2011 2:25 am
prateek_guy2004 wrote:
[email protected] wrote:The probability of having a girl in all 3 cases is :
=> 1/2*1/2*1/2= 1/8

The probability of NOT having a boy in all 3 cases is :
=> 1/2*1/2*1/2= 1/8

Therefore the probability of having a girl and not a boy will be;

=> 1/8 +1/8
=> 2/8
=> 1/4

Ans : 1/4
This is a easy approach.....
I feel this is not answering the question correctly. Either all boy or all girl. It answers all girls and not a boy.

Somebody clarify.
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by JS_2 » Mon Sep 12, 2011 10:03 pm
1/4
2 favourable outcomes: either all boys or all girls
All possible outcomes: (either boy or girl=2)*(either boy or girl=2)*(either boy or girl=2)

i.e, 2/2*2*2= 2/8=1/4.
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by prashant misra » Thu Sep 15, 2011 7:00 am
thank you stuart sir for explaining different approach towards the question.The answer is D
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by vini1612 » Tue Sep 20, 2011 8:32 am
niazsna786 wrote:The question says, the probability of all the 3 children be same gender

The children can be all 3 girls or all 3 boys.

The probability of getting a boy = 1/2
The probability of getting a girls = 1/2


Probability of getting 3 girls = 1/2 * 1/2 * 1/2 = 1/8
Probability of getting 3 boys= 1/2 * 1/2 * 1/2 = 1/8

What is asked is probability of getting of same gender i.e. either all 3 girls or all 3 boys i.e. 1/8 + 1/8 = 1/4
E is the answer
Thanks! this made me understand the solution easily!
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by parul9 » Fri Oct 07, 2011 11:56 pm
E - 1/4!
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by nsamkari » Sat Oct 22, 2011 11:35 pm
The answer is 1/4
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by Deependra1 » Wed Oct 26, 2011 4:40 am
Answer: A
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by Sharma_Gaurav » Mon Jan 09, 2012 3:38 pm
straight E is the answer
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by mparakala » Wed Nov 07, 2012 8:01 am
E

NS: 2/8 = 1/4
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by rajeshsinghgmat » Mon Jan 28, 2013 12:07 am
sEE the ansEr.

P(E)= (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2)
= (1/2)^3 + (1/2)^3
= 1/8 +1/8
= 2/8
= 1/4
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by Mathsbuddy » Thu Nov 28, 2013 12:35 am
P(1st child male or female) = 1
P(2nd child matching previous) = 1/2
P(3rd child matching previous) = 1/2

Total probability = 1 * 1/2 * 1/2 = 1/4
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by lulufrenchie » Wed Apr 09, 2014 11:21 pm
Hi all,

I draw a probability tree to answer this question and found the right answer.

But is it a waist of time ?
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by Brent@GMATPrepNow » Thu Apr 10, 2014 8:08 am
lulufrenchie wrote:Hi all,

I draw a probability tree to answer this question and found the right answer.

But is it a waist of time ?
A probability tree doesn't take long AND it provides an easy way to SEE all of the possible outcomes:
Image

Of the 8 possible outcomes, let's highlight those where all 3 children are the same gender.
Image

There are 2 such outcomes, so the probability = 2/8 = [spoiler]1/4 = E[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by bml1105 » Tue May 20, 2014 6:46 pm
I know this is an old thread, but what's the difference between the original question and this question:

"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 2 boys?"

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16


In the problem I posted, the solution is to use permutation/combination instead of multiplying the fractions
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by [email protected] » Tue May 20, 2014 11:12 pm
Hi bml1105,

Your question can also be solved in a variety of ways. Since there are 4 children and each is either a boy or a girl, then you could draw a probability tree (there are 2^4 = 16 possibilities) or you could use permutations/combinations or you could use probability math. Most questions of this type can be solved in more than 1 way, so the measure of your approach is if it gets you the correct answer AND is relatively fast to use.

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