## Probability

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### Probability

by Dream Weaver » Tue Jul 21, 2009 11:16 am
. The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?

a) 1/8.
b) 1/6.
c) 1/3.
d) 1/5.
e) 1/4.

OA is E

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by niharikamohanty@gmail.com » Tue Jul 21, 2009 11:31 am
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

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### Ok..

by Dream Weaver » Tue Jul 21, 2009 11:34 am
Well, I myself arrived at the OA in a diifernt way..

The gender of the first-born is insignificant since we want all children to be of the same gender no matter if they are all boys or girls.
The probability for the second child to be of the same gender as the first is: ½. The same probability goes for the third child. Therefore the answer is ½ x ½ = ¼.

But, I want to see other approaches to this problem..

Thanks !

DW

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### Re: Probability

by Stuart@KaplanGMAT » Tue Jul 21, 2009 11:54 am
Dream Weaver wrote:. The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?

a) 1/8.
b) 1/6.
c) 1/3.
d) 1/5.
e) 1/4.
Both approaches noted are great and, in a simple question like this, effective ways to solve the problem.

We could also recognize that this is a pseudo coin-flip question. Any time you see 50/50 binary situations, think coin flips!

The question could have been:

if a fair coin is flipped 3 times, what's the probability of getting either exactly 3 heads or exactly 3 tails?

So, we could have applied the coin flip formula, which is:

Probability(k results out of n flips) = nCk/2^n

Since we want 3 heads OR 3 tails, we calculate the probability of each and ADD them together:

3C3/2^3 + 3C3/2^3 = 1/8 + 1/8 = 1/4

You also could have brute forced the question, much like you can with 3 coin flip questions. The possibilities are:

BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG

and 2 out of the 8 match what we desire, so 2/8 or 1/4 is the right answer.

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### Thanks

by Dream Weaver » Tue Jul 21, 2009 12:02 pm
Hi,
Thanks a lot Stuart for the concept !!

This is exactly why I posted this question here..

DW

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by bacchewar_prashant » Thu Nov 18, 2010 6:00 am
I used following approach. Let us say I have three letters with me and I want to arrange them.

1. All the letters are same e.g. BBB or GGG then there is only one way of arranging them.
2. Now if I have 2Bs and one G then I have three options
BBG
GBG
GBB
3. If I have 1B and two Gs so again I have three options.
BGG
GBG
GGB

Thus I can have 8 possible options
BBB
GGG
BBG
GBG
GBB
BGG
GBG
GGB

Now consider this in context of problem assume the alphabet B is boy and alphabet G is girl

We have total 8 options as listed above
Out of these 8 options two options BBB and GGG are cases where children are of same gender.
so P = no of outcomes of event/Total no. of outcomes
= 2/8 = 1/4

I am not sure whether I am clear. Please correct if something is wrong in my logic

Thanks

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by oldguy » Mon Apr 04, 2011 6:47 am
can't one just calculate p(BBB) = P(B1*P(B2)*P(B3) = 1/2 * 1/2 * 1/2 = 1/8

p(GGG) = P(G1)*P(G2)*P(G3) = 1/2 * 1/2 * 1/2 = 1/8

p(BBB)+ p(GGG)= 1/8 + 1/8 = 1/4

anything wrong with this straightforward approach?

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by Geva@EconomistGMAT » Mon Apr 04, 2011 7:24 am
It's fine, as long as you indeed remember that there are two scenarios that need to be added together.
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by niazsna786 » Mon Apr 04, 2011 6:42 pm
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

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by sushantgupta » Sun May 01, 2011 10:59 pm
Probability of having 3 girls = 1/2*1/2*1/2 = 1/8
Probability of having 3 boys = 1/2*1/2*1/2 = 1/8

Pobability of having same gender = 1/8+1/8 = 1/4

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by GMATGuruNY » Mon May 02, 2011 7:02 am
Dream Weaver wrote:. The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?

a) 1/8.
b) 1/6.
c) 1/3.
d) 1/5.
e) 1/4.

OA is E
The gender of the first child is irrelevant: the first child can be either a boy or a girl.
We care only that the second child and the third child are of the same gender as the first child.

P(2nd child is the same gender) = 1/2.
P(3rd child is the same gender) = 1/2.

Since we want both outcomes to happen, we multiply the fractions:
1/2 * 1/2 = 1/4.

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by MBA.Aspirant » Sat Jul 30, 2011 5:08 am
Can you just say probability of all same gender = (1/2)^3 and this can happen in 2 ways (all boys or all girls) so 2 *1/8 = 1/4. Is this approach correct?

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by olegpoi » Thu Sep 01, 2011 12:27 pm
IMO A
(1/2)^3

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by prateek_guy2004 » Thu Sep 01, 2011 1:09 pm
niharikamohanty@gmail.com 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.....
Don't look for the incorrect things that you have done rather look for remedies....

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by saketk » Thu Sep 01, 2011 8:26 pm
This question is similar to a coin which is flipped 3 times. As we know the probability of a head and a tail is same ( in
This question boys and girls), we can calculate the total outcomes - 2*2*2 =8
also the chances of having all boys or all girls is one each. So total favorable outcome =2
probability = 2/8=1/4
option E is the right answer