Last visit was: 21 Jul 2024, 20:43 It is currently 21 Jul 2024, 20:43
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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Manager
Manager
Joined: 06 Dec 2012
Status:struggling with GMAT
Posts: 97
Own Kudos [?]: 1565 [39]
Given Kudos: 46
Location: Bangladesh
Concentration: Accounting
GMAT Date: 04-06-2013
GPA: 3.65
Send PM
Most Helpful Reply
User avatar
Manager
Manager
Joined: 31 May 2012
Posts: 101
Own Kudos [?]: 411 [20]
Given Kudos: 69
Send PM
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 19175
Own Kudos [?]: 22687 [7]
Given Kudos: 286
Location: United States (CA)
Send PM
General Discussion
avatar
Intern
Intern
Joined: 17 Nov 2012
Posts: 11
Own Kudos [?]: 8 [2]
Given Kudos: 7
Send PM
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
2
Kudos
I would present another approach.

P(z odd) = P(boys odd)* P(girls even) + P(boys even)* P(girls odd)
= 2/C2,3 * 1/C2,3 + 1/C2,3 * 2/C2,3
= 4/9

P(z even) = 1 - P(z odd) = 5/9

|P(z even)-P(z odd)| = 1/9
Senior Manager
Senior Manager
Joined: 02 Dec 2014
Posts: 304
Own Kudos [?]: 305 [1]
Given Kudos: 353
Location: Russian Federation
Concentration: General Management, Economics
GMAT 1: 640 Q44 V33
WE:Sales (Telecommunications)
Send PM
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
1
Kudos
umeshpatil wrote:
Age of Boys:4, 6, 7
Sum of ages taken 2 at a time: 10,13,11


Ages of Girls:5, 8, 9
Sum of ages taken 2 at a time: 13,17,14


9 Combinations of sum between sets(10,12,11) & (13,17,14)
=23,27,24- 16,30,17- 24,28,25

Prob(Even)= 5/9
Prob(Odd) =4/9

Answer=5/9 - 4/9 = 1/9

I think 13 should be here not 12. And i don't understand second bolded fragment. May be this should be (26,30,27)?
User avatar
Manager
Manager
Joined: 04 May 2015
Posts: 64
Own Kudos [?]: 30 [2]
Given Kudos: 58
Concentration: Strategy, Operations
WE:Operations (Military & Defense)
Send PM
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
2
Bookmarks
Please feel free to critique me here but this is how I solved.

Boys: 4, 6, 7
Girls: 5, 8, 9
Number of possible combinations of boys or girls = 3!/2! = 3
i.e. there is 3 possible combinations of girls and 3 of boys

Probability that sum of 2 boys ages is even = 1/3 [a]
Probability that sum of 2 boys ages is odd = 2/3 [b]
Probability that sum of 2 girls ages is even = 1/3 [c]
Probability that sum of 2 girls ages is odd = 2/3 [d]

probability that sum of 2 girls and 2 boys is even = [a]*[c] + [b]*[d] = 5/9 [e]
probability that sum of 2 girls and 2 boys is odd = [a]*[b] + [c]*[d] = 4/9 [f]

Therefore the differences in the probabilities is [e] - [f] = 1/9

No idea if this is correct or if it was dumb luck. I am struggling daily with this GMAT journey so would appreciate the feedback. :)
Director
Director
Joined: 26 Oct 2016
Posts: 506
Own Kudos [?]: 3411 [1]
Given Kudos: 877
Location: United States
Concentration: Marketing, International Business
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE:Education (Education)
Send PM
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
1
Kudos
Total ways to choose the 4 children:
Number of pairs of boys that can be formed from 3 options = 3C2 = 3.
Number of pairs of girls that can be formed from 3 options = 3C2 = 3.
To combine these options, we multiply:
3*3 = 9.

We could list all the ways to pick four children—two boys and two girls. We could also take the opposite approach: list all the ways to leave out two children—one boy and one girl. There are fewer scenarios to list with the left-out children, so let's take that approach. The sum of the ages of all six children is (4 + 6 + 7) + (5 + 8 + 9) = 39, an odd number. We can then list all 9 scenarios, subtracting out the ages of the left-out children.

Of the 9 scenarios listed, 5 yield an even z and 4 yield an odd z. Each outcome is equally likely.
The difference between the probability that z is even and the probability that z is odd is therefore 5/9 – 4/9 = 1/9.
The correct answer is A

Table pasted below to represent all 9 scenarios.
Attachments

table.PNG
table.PNG [ 263.31 KiB | Viewed 8932 times ]

Manager
Manager
Joined: 20 Mar 2019
Posts: 146
Own Kudos [?]: 15 [0]
Given Kudos: 282
Location: India
Send PM
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
Nice question. I did it the following way:

Pairs for when z is odd
4,5 4,9 6,5 6,9 7,8
Each pair has probability of 1/3
So 1/3 * 1/3 * 1/3 ...5 times (5 pairs) = 5/9

Pairs for when z is even
4,8 6,8 6,5 6,9 7,8

Again each pair has probability of 1/3 ... 4 pairs = 4/9

5/9 - 4/9 = 1/9

Answer A

Bunuel I tried to do this question through combination but wasn't able to. Any idea how to do that? TIA
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 34040
Own Kudos [?]: 853 [0]
Given Kudos: 0
Send PM
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
Moderator:
Math Expert
94441 posts