Probability

Expert replies

arorag: Master | Next Rank: 500 Posts; Posts: 343; Joined: Mon Jan 21, 2008 3:28 pm; Thanked: 4 times

by arorag » Sun Jun 29, 2008 5:36 pm

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
ANS[spoiler] 4/9[/spoiler]

Like this post

Source: Beat The GMAT — Problem Solving | Sun Jun 29, 2008 5:36 pm

punit.kaur.mba: Master | Next Rank: 500 Posts; Posts: 104; Joined: Wed Feb 13, 2008 1:44 pm; Thanked: 8 times

by punit.kaur.mba » Tue Jul 01, 2008 8:10 am

According to me the answer should be 9/15 = 3/5

4/9 looks wrong to me. Reason being, he is simple taking 2/3 * 2/3 = 4/9

Doesnt make sense...

Here is how I get 3/5 -
Denominator - Total no of ways to get 4 children = 6C4
Numerator - Total no of ways to select 2 girls of 3 * Total no of ways to select 2 boys of 3
=(3C2 * 3C2)/6C4
=9/15 = 3/5

arorag, could you please check again. I doubt the OA. Whats the source of the quest..

Anybody has any other explainations??

https://punitkaur.blocked/

Like this post

GMAT/MBA Expert

Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Tue Jul 01, 2008 10:51 am

Yes, the OA (4/9) can't be right here. The answer is 3/5; I used the same approach as in the above post.

Like this post

beeparoo: Master | Next Rank: 500 Posts; Posts: 200; Joined: Sun Jun 17, 2007 10:46 am; Location: Canada; Thanked: 9 times

by beeparoo » Wed Jul 02, 2008 5:28 pm

Ian Stewart wrote:Yes, the OA (4/9) can't be right here. The answer is 3/5; I used the same approach as in the above post.

Thanks for saying something, punit, Ian! I was concerned too when I saw the answer. Phew.

Like this post

somail: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Tue Jul 01, 2008 11:54 am; Location: San Diego

by somail » Thu Jul 03, 2008 10:57 am

Hi,

For some reason I am having a hard time understanding the probablity questions that are usually posted. Can one of you solve this in a little more detail (dumbed down).

Thanks for any help

Like this post

pranavc: Master | Next Rank: 500 Posts; Posts: 167; Joined: Tue Feb 26, 2008 4:37 amFollowed by: 1 member; Thanked: 5 times

by pranavc » Thu Jul 03, 2008 6:38 pm

I'll give this a shot. You have three guys and three girls to select from. You have 4 spots. The order in these 4 spots is insignificant. So you have to use the formula of combinations to determine the total number of selection possibilities. This boils down to 6C4 (i.e. six people, 4 spots, and order does not matter). This gives you a total of 15 possibilities.

Now, think of the number of ways in which the number of girls will NOT be equal to the number of guys. Two scenarios will results in this.

Scenario 1: 3 girls are selected and one boy is selected

The number of selections that would fit this scenario = 3

because one of three boys can be selected as the "one boy with the three girls".

Scenario 2: 3 boys are selected and one girl is selected

The number of selections that would fit this scenario = 3

because one of three girls can be selected as the "one girl with the three boys".

Hence, the total number of selections in which the number of boys is not equal to the number of girls is 6.

Hence, the number of selections in which the number of boys IS equal to the nunber of girls = 15 (i.e. total number of selection possibilities) - 6

= 9

Hence, the probability of this equal situation = 9/15

I hope this makes sense.

Like this post

somail: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Tue Jul 01, 2008 11:54 am; Location: San Diego

by somail » Mon Jul 07, 2008 12:22 pm

Thanks, that made a lot of sense. My biggest problem was that I could not figure out how you guys were getting 15. After you mentioned the "formula of combinations", I googled it and found the equation. The problem was much easier after that.

Like this post

ccissilly: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Sun Nov 10, 2013 3:31 pm

by ccissilly » Sun Nov 10, 2013 3:33 pm

ok, isnt 6C4 equal to 30? 6*5*4*3*2*1/4*3*2*1?

Like this post

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Sun Nov 10, 2013 10:24 pm

Hi ccissilly,

The Combination Formula is:

N! / [K!(N-K)!]

So 6c4 = 6! / [4!(6-4)!] = 6!/4!2! = 15

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

Like this post

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Sun Nov 10, 2013 11:13 pm

3B 3G --> 4 to be selected

(15 - 3B3*3G1 - 3G3*3B1) = 15 - 1*3 - 1*3 = 9

6c4 = 6x5/2 = 15

Probability = 9/15 = 3/5

R A H U L

Like this post

pareekbharat86: Master | Next Rank: 500 Posts; Posts: 168; Joined: Thu Nov 01, 2012 7:43 pm; Thanked: 1 times

by pareekbharat86 » Mon Nov 11, 2013 9:00 am

Out of 3 girls, choose 2: 3C2= 3

Out of 3 boys, choose 2: 3C2= 3

Therefore, total ways= 3*3 =9

Total ways of choosing 4 persons out of 6= 6C4= 15

Therefore, total probability= 9/15= 3/5

Thanks,
Bharat.

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 » Mon Nov 11, 2013 9:34 am

Another brute force way is:
We can have a group of 4 either 1B3G, or 2B2G, or 3B1G.
Case1: 3B1G = B1B2B3G1 or B1B2B3G2 OR B1B2B3G3 = total 3 ways
Case3: 3G1B = similar to above = total 3 ways
Case2: 2B2G = {B1B2G1G2, B1B2G2G3, B1B2G1G3} now replace - B1B2 with B2B3 and afterwards with B3B1 (with same G* combinations).
Total combinations will be 3+3+3 = 9

Case 2 is favourable combination: probability = 9/(3+3+9)=9/15=3/5.
As mentioned this is brute force and cannot come in handy with complex combinations.

a lil' Thank note goes a long way

Like this post

stephny: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Mon Nov 11, 2013 1:02 am

by stephny » Tue Nov 12, 2013 12:58 am

please break down your calculation of 6C4 and how that equals 15

Like this post

stephny: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Mon Nov 11, 2013 1:02 am

by stephny » Tue Nov 12, 2013 12:59 am

please break down your calculation of 6C4 and how that equals 15

Like this post

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Tue Nov 12, 2013 1:02 am

stephny wrote:please break down your calculation of 6C4 and how that equals 15

Combination formula : nCr ==> n!/r!(n-r)!

So, 6C4 ==> 6!/4!(6-4)! ==> 6!/4!2! ==> 6x5x4!/4!/2! ==> 6x5/2x1 ==> 3x5 ==> 15

I hope it's clear now!

R A H U L

Like this post

Post new topic Post Reply

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