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Senior Manager
Status: struggling with GMAT
Joined: 06 Dec 2012
Posts: 323
Location: Bangladesh
Concentration: Accounting
GMAT Date: 04-06-2013
GPA: 3.65
Followers: 5
Kudos [?]:
28
[0], given: 46
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Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
 09 Dec 2012, 10:05
Question Stats:
71% (02:15) correct
28% (01:35) wrong based on 1 sessions
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
Last edited by Bunuel on 09 Dec 2012, 10:09, edited 2 times in total.
Renamed the topic and edited the question.
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Manager
Joined: 31 May 2012
Posts: 123
Followers: 1
Kudos [?]:
23
[2] , given: 48
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
09 Dec 2012, 10:27
2
This post received
KUDOS
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
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Intern
Joined: 17 Nov 2012
Posts: 20
Followers: 1
Kudos [?]:
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[0], given: 6
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
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
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
[#permalink]
01 Jan 2013, 15:19
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