GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Nov 2018, 09:25

Gmatbusters' Weekly Quant Quiz

Join here 


Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • FREE Quant Workshop by e-GMAT!

     November 18, 2018

     November 18, 2018

     07:00 AM PST

     09:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score. November 18th, 7 AM PST
  • How to QUICKLY Solve GMAT Questions - GMAT Club Chat

     November 20, 2018

     November 20, 2018

     09:00 AM PST

     10:00 AM PST

    The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.

Ms. Barton has four children. You are told correctly that she has at

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
B
Joined: 04 Apr 2017
Posts: 19
Re: Ms. Barton has four children. You are told correctly that she has at  [#permalink]

Show Tags

New post 20 Jun 2018, 07:11
Since there is at least 2 girls, then we have 3 probabilities:
1) 2 boys & 2 girls:
4!/(2!*2!) = 6

2) 1 boy & 3 girls:
4!/3! = 4

3) 4 girls: 1

So the total probabilities are 6+4+1=11
We need the probability of 2B & 2G = 6/11
Director
Director
User avatar
P
Joined: 14 Dec 2017
Posts: 508
Premium Member
Re: Ms. Barton has four children. You are told correctly that she has at  [#permalink]

Show Tags

New post 20 Jun 2018, 10:09
Onell wrote:
Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)

(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11


Ms. Barton has at least 2 girls out the 4 children. We need probability of 2 boys & 2 girls.

We have three cases: GGBB, GGGB & GGGG

GGBB can occur in 4!/2!2! = 6 ways

GGGB can occur in 4!/3! = 4 ways

GGGG can occur in only 1 way

Total # of ways = 6 + 4 + 1 = 11

Probability of GGBB = 6/11

Answer E.


Thanks,
GyM
_________________

New to GMAT Club - https://gmatclub.com/forum/new-to-gmat-club-need-help-271131.html#p2098335

Senior Manager
Senior Manager
avatar
S
Joined: 04 Aug 2010
Posts: 306
Schools: Dartmouth College
Re: Ms. Barton has four children. You are told correctly that she has at  [#permalink]

Show Tags

New post 20 Jun 2018, 11:07
1
dprchem wrote:
Why are we getting into arrangements. The question is about probability (no matter in which order). Then the answer should be 1/4. why is it wrong?


Ms. Barton currently has two girls. If the probability of having a girl is equal to the probability of having a boy, what is the probability that her next two children are boys?

Here, the first 2 children must be girls.
Below are all the ways for the first 2 children to be girls:
GGBB
GGBG
GGGB
GGGG
Of the 4 possible outcomes, only the 1 case in blue includes 2 boys.
Thus:
P(the next 2 children are boys) = 1/4.

The posted problem is different.
It asks the following:

Of Ms. Barton's 4 children, at least two are girls. If the probability of having a girl is equal to the probability of having a boy, what is the probability that there are 2 boys among Ms. Barton's 4 children?

Here, the first two children do NOT have to be girls.
The only condition is that AT LEAST 2 of the 4 children must be girls.
As a result, a greater number of outcomes are possible.
Below are all the ways for at least 2 girls to be among the 4 children:

4 girls:
GGGG

3 girls:
GGGB
GGBG
GBGG
BGGG

2 girls:
GGBB
GBGB
GBBG
BGGB
BGBG
BBGG
BBGG


Of the 11 ways to have at least 2 girls, only the 6 cases in blue include 2 boys.
Thus:
P(2 boys are among the 4 children) = 6/11.


_________________

GMAT and GRE Tutor
Over 1800 followers
Click here to learn more
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.

Intern
Intern
avatar
B
Joined: 29 Aug 2017
Posts: 30
Location: India
Concentration: General Management, Leadership
GMAT 1: 640 Q49 V27
GPA: 4
Reviews Badge
Re: Ms. Barton has four children. You are told correctly that she has at  [#permalink]

Show Tags

New post 30 Jun 2018, 23:00
dprchem wrote:
VeritasPrepKarishma wrote:
forevertfc wrote:
What am I missing here?

The questions clearly states that 2 of the 4 kids are girls. It also clearly asks "what is the probability that the other 2 are boys?"

This gives us a P of 1/2 that a child will be a boy and 1/2 that the child will be a girl. Armed with the fact that having two girls will have no significance on the sex of the other two children, we can treat this as mutually exclusive scenario.

So there is P:1/2 that one of the remaining children is a boy, and P:1/2 that the other child is also a boy. Therefore, the probability would become 1/4.

Am I reading this question wrong?


We are concerned about the 4 selections together, not the individual selection. What this means is that if your logic is correct, the probability of having 4 girls would be the same as the probability of having 2 girls and 2 boys. But is that correct? No it is not.

In how many ways can you have 4 girls?
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next girl, next girl, next girl)

In how many ways can you have 2 girls and 2 boys?
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next girl, next boy, next boy)
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next boy, next girl, next boy)
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next boy, next boy, next girl)
and so on...

There is a sequence to obtaining kids and that needs to be factored in. Many ways will lead to 2 girls and 2 boys hence the probability of obtaining 2 girls and 2 boys will be higher than the probability of obtaining 4 girls. The question clearly tells us that we don't know which 2 are girls so we cannot say that this the case where first child and second child are girls and others are boys. Hence you need to follow the method discussed above: http://gmatclub.com/forum/ms-barton-has ... 58#p991399

It is similar to the coin flipping case. Every coin flip is independent of the other but the probability of obtaining 4 heads in 4 flips is less than the probability of obtaining 2 heads and 2 tails.


Why are we getting into arrangements. The question is about probability (no matter in which order). Then the answer should be 1/4. why is it wrong?


It is wrong because in Your solution You are considering that there are 2 girls. But according to question there are at least 2 girls.
So that why we need to make cases and solve the question.
Intern
Intern
avatar
B
Joined: 10 Mar 2018
Posts: 12
Re: Ms. Barton has four children. You are told correctly that she has at  [#permalink]

Show Tags

New post 09 Jul 2018, 14:13
VeritasPrepKarishma wrote:
raghupara wrote:
Hi people,

Suddenly, something comes to my mind so awfully, but I couldn't disprove why I am wrong. Could someone help me in this.

I suddenly see it as : P(2 girls, 2 boys) = 1/2*1/2*1/2*1/2 = 1/16.(since each has a prob of 1/2).

Pls disprove me.... I don see where I am wrong, but I know surely I am.

Thanks


The following is the complete explanation for this question:

You need to find Probability of 2 boys and 2 girls given there are at least 2 girls. This is a conditional probability. You have already been given that there are at least 2 girls. You need to find the probability of 2 boys and 2 girls given this condition.
P(A given B) = P(A)/P(B)

P(2 boys, 2 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8
You multiply by 4!/(2!*2!) because out of the four children, any 2 could be boys and the other two would be girls. So you have to account for all arrangements: BGBG, BBGG, BGGB etc

P(atleast 2 girls) = P(2 boys, 2 girls) + P(1 boy, 3 girls) + P(4 girls) = 1 - P(4 boys) - P(3 boys, 1 girl)
You can solve P(atleast 2 girls) in any way you like.
P(1 boy, 3 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4
P(4 girls) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16
P(atleast 2 girls) = 3/8 + 1/4 + 1/16 = 11/16

P(2 boys, 2 girls given atleast 2 girls) = P(2 boys, 2 girls) / P(atleast 2 girls) = (3/8)/(11/16) = 6/11

In the above given solutions, all the (1/2)s have been ignored because the probability of a boy is 1/2 and of a girl is 1/2 too. Hence in every case you will get (1/2)*(1/2)*(1/2)*(1/2) which can be ignored. We just need to focus on arrangements in each case. This is similar to the famous coin flipping questions (the probability of 3 Heads and a tail etc) (check out Binomial Probability in Veritas Combinatorics book for an explanation)


Hi @VeritasPrepKrishna

I am not able to understand why arrangement matter here?

My solution:

there can be either 2 girls, 3 girls or 4 girls i.e. 3 condition. The only possibility for two boys is when family has two girls and two boys i.e. one condition

so probability will be 1/3.

Can you please correct my understanding.
GMAT Club Bot
Re: Ms. Barton has four children. You are told correctly that she has at &nbs [#permalink] 09 Jul 2018, 14:13

Go to page   Previous    1   2   [ 25 posts ] 

Display posts from previous: Sort by

Ms. Barton has four children. You are told correctly that she has at

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.