Author 
Message 
TAGS:

Hide Tags

Manager
Status: struggling with GMAT
Joined: 06 Dec 2012
Posts: 117
Location: Bangladesh
Concentration: Accounting
GMAT Date: 04062013
GPA: 3.65

Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
Updated on: 09 Dec 2012, 10:09
Question Stats:
68% (02:52) correct 32% (02:49) wrong based on 236 sessions
HideShow timer Statistics
Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd? (A) 1/9 (B) 1/6 (C) 2/9 (D) 1/4 (E) 1/2
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by mun23 on 09 Dec 2012, 10:05.
Last edited by Bunuel on 09 Dec 2012, 10:09, edited 2 times in total.
Renamed the topic and edited the question.




Manager
Joined: 31 May 2012
Posts: 108

Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
09 Dec 2012, 10:27
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




Intern
Joined: 17 Nov 2012
Posts: 17

Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
01 Jan 2013, 15:19
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
Joined: 02 Dec 2014
Posts: 352
Location: Russian Federation
Concentration: General Management, Economics
WE: Sales (Telecommunications)

Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
23 Apr 2015, 14:32
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)?
_________________
"Are you gangsters?"  "No we are Russians!"



Manager
Joined: 04 May 2015
Posts: 69
Concentration: Strategy, Operations
WE: Operations (Military & Defense)

Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
13 Jul 2015, 12:06
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
Joined: 26 Oct 2016
Posts: 599
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
13 Mar 2017, 01:48
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 leftout 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 leftout 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 [ 263.31 KiB  Viewed 2661 times ]
_________________
Thanks & Regards, Anaira Mitch



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9032
Location: United States (CA)

Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
12 Sep 2018, 18:36
mun23 wrote: Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd?
(A) 1/9 (B) 1/6 (C) 2/9 (D) 1/4 (E) 1/2 The sum of the two selected boys can be either 4 + 6 = 10, 4 + 7 = 11 or 6 + 7 = 13. Thus, there is a 1/3 probability that the sum of the ages of the two boys will be even and 2/3 probability that the sum of the ages of the two boys will be odd. Similarly, the sum of the two selected girls can be either 5 + 8 = 13, 5 + 9 = 14 or 8 + 9 = 17. Thus, there is a 1/3 probability that the sum of the ages of the two girls will be even and 2/3 probability that the sum of the ages of the two girls will be odd. Now, let’s first find the probability that z is even. Since z is the sum of ages of the selected boys and girls, z can be even if the sum of both the selected boys and selected girls ages are even or if sum of both the selected boys and selected girls ages are odd. The probability that both sums are even is 1/3 x 1/3 = 1/9 and the probability that both sums are odd is 2/3 x 2/3 = 4/9. Thus, there is a 1/9 + 4/9 = 5/9 probability that z is even. Similarly, let’s find the probability that z is odd. z can only be odd of one of the sums is even and the other is odd. Probability that the boys sum is even and girls sum is odd is 1/3 x 2/3 = 2/9. Probability that the boys sum is odd and the girls sum is even is 2/3 x 1/3 = 2/9. Thus, there is a 2/9 + 2/9 = 4/9 probability that z is odd. Finally, the difference between the probabilities that z is even and z is odd is 5/9  4/9 = 1/9. Answer: A
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



NonHuman User
Joined: 09 Sep 2013
Posts: 13963

Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
Show Tags
08 Dec 2019, 10:32
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.
_________________




Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
08 Dec 2019, 10:32






