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OG quant probability

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by Redhorsep » Fri Sep 23, 2011 8:47 am
Hi,

Please help with this problem from OG quant second edition. It would be great if there's an alternative way to do the problem besides listing the possibibilities. Thanks!


***
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?
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by cans » Fri Sep 23, 2011 9:08 am
select 2: 4C2. they are girls: (1/2)(1/2)
other 2 are boys: (1/2)(1/2)
thus 4C2*1/16 = 3/13
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by gmatclubmember » Fri Sep 23, 2011 9:11 am
The total outcomes possible are: 4B,3B1G,2B2G,1B3G,4G.
So the probability of having 2B2G is 1/5

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by knight247 » Fri Sep 23, 2011 9:28 am
Each child could be either a boy or a girl i.e. either outcome has a 50-50 chance or 1/2 probability

Now, the outcomes need to be BBGG and the different permutations of it. 4!/(2!2!)=6

Since, its a binomial probability(where either outcome has an equal chance) we multiply the different probabilities as 1/2*1/2*1/2*1/2=1/16

And, now multiplying the two we have 6*1/16=6/16=[spoiler]3/8[/spoiler]
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by Redhorsep » Fri Sep 23, 2011 9:42 am
the correct answer is 3/8
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by mehrasa » Fri Sep 23, 2011 9:46 am
p(2 boys and 2 girls)= i/2 ^4= 1/16
also we have to take into account the ways of their arrangement ;), bcuz order is important ==>(2B,2G)= 4C2=6

==>final probability is 6 *1/16=6/16
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by Redhorsep » Fri Sep 23, 2011 9:47 am
knight247 wrote:Each child could be either a boy or a girl i.e. either outcome has a 50-50 chance or 1/2 probability

Now, the outcomes need to be BBGG and the different permutations of it. 4!/(2!2!)=6

Since, its a binomial probability(where either outcome has an equal chance) we multiply the different probabilities as 1/2*1/2*1/2*1/2=1/16

And, now multiplying the two we have 6*1/16=6/16=[spoiler]3/8[/spoiler][/quote

Can you explain the logic behind the permutation formula you came up with?
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by gmatclubmember » Fri Sep 23, 2011 9:53 am
gmatclubmember wrote:The total outcomes possible are: 4B,3B1G,2B2G,1B3G,4G.
So the probability of having 2B2G is 1/5

Cheers
Ami/-
Looks like I got this TOTALLY incorrect.
Could someone please explain why the above reasoning is incorrect, I am still not able to comprehend the flaw in my reasoning?

Cheers
Ami/-
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by Redhorsep » Fri Sep 23, 2011 9:57 am
mehrasa wrote:p(2 boys and 2 girls)= i/2 ^4= 1/16
also we have to take into account the ways of their arrangement ;), bcuz order is important ==>(2B,2G)= 4C2=6

==>final probability is 6 *1/16=6/16
can you explain where does it say in the problem that order matters?
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by knight247 » Fri Sep 23, 2011 10:10 am
@gmatclubmember
U've considered the unordered pairs only. The options u've listed are correct if the order didn't matter.For example under 2B2G, BBGG is different from BGBG and from BGGB etc

@redhorse
You can infer from the above that order is important. Look thru my post where I have arranged BBGG in 4!/2!2! ways which is 4! divided by (the number of times B is repeated*the number of time G is repeated)..Hope this clarifies ur doubt
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by shekhar.kataria » Sat Sep 24, 2011 10:31 pm
knight247 wrote:@gmatclubmember
U've considered the unordered pairs only. The options u've listed are correct if the order didn't matter.For example under 2B2G, BBGG is different from BGBG and from BGGB etc

@redhorse
You can infer from the above that order is important. Look thru my post where I have arranged BBGG in 4!/2!2! ways which is 4! divided by (the number of times B is repeated*the number of time G is repeated)..Hope this clarifies ur doubt


So you mean bcz of having equal probabilities in both boy and girl case above. We will need permutations here.
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by dhonu121 » Sun Sep 25, 2011 6:53 am
knight247 wrote:@gmatclubmember
U've considered the unordered pairs only. The options u've listed are correct if the order didn't matter.For example under 2B2G, BBGG is different from BGBG and from BGGB etc

@redhorse
You can infer from the above that order is important. Look thru my post where I have arranged BBGG in 4!/2!2! ways which is 4! divided by (the number of times B is repeated*the number of time G is repeated)..Hope this clarifies ur doubt
The question asked the probability that the couple has 2 boys and 2 girls.How can we be sure that order should be taken into account here ?

I find the answer posted above as 1/5 correct since the question just asks the probability that the couple has 2 boys and 2 girls.Not matter what their order is.

Can someone please elaborate on this.
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