OG quant probability

Expert replies

Redhorsep: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Thu Jun 30, 2011 2:47 pm; Thanked: 1 times

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?

Like this post

Source: Beat The GMAT — Problem Solving | Fri Sep 23, 2011 8:47 am

cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

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

If my post helped you- let me know by pushing the thanks button

Contact me about long distance tutoring!
[email protected]

Cans!!

Like this post

gmatclubmember: Master | Next Rank: 500 Posts; Posts: 158; Joined: Sat Sep 03, 2011 10:31 amFollowed by: 2 members; Thanked: 29 times

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

Cheers
Ami/-

Like this post

knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

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]

Like this post

Redhorsep: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Thu Jun 30, 2011 2:47 pm; Thanked: 1 times

by Redhorsep » Fri Sep 23, 2011 9:42 am

the correct answer is 3/8

Like this post

mehrasa: Master | Next Rank: 500 Posts; Posts: 279; Joined: Fri Nov 05, 2010 5:43 pmFollowed by: 1 member; Thanked: 15 times

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

Like this post

Redhorsep: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Thu Jun 30, 2011 2:47 pm; Thanked: 1 times

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?

Like this post

gmatclubmember: Master | Next Rank: 500 Posts; Posts: 158; Joined: Sat Sep 03, 2011 10:31 amFollowed by: 2 members; Thanked: 29 times

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/-

Like this post

Redhorsep: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Thu Jun 30, 2011 2:47 pm; Thanked: 1 times

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?

Like this post

knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

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

Like this post

shekhar.kataria: Master | Next Rank: 500 Posts; Posts: 268; Joined: Mon Mar 15, 2010 12:10 am; Thanked: 13 times

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.

Restlessness and discontent are the first necessities of progress.--Thomas A. Edison

If you find this post helpful, let me know by clicking thanks above

Like this post

dhonu121: Master | Next Rank: 500 Posts; Posts: 316; Joined: Sun Aug 21, 2011 6:18 amFollowed by: 6 members; Thanked: 16 times

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.

If you've liked my post, let me know by pressing the thanks button.

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review