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Three boys are ages 4, 6 and 7 respectively. Three girls are

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Senior Manager
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Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink] New post 09 Dec 2012, 09:05
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Question Stats:

68% (02:54) correct 32% (01:39) wrong based on 63 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Dec 2012, 09:09, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink] New post 09 Dec 2012, 09:27
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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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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink] New post 01 Jan 2013, 14:19
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] New post 09 Jul 2014, 06:00
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are   [#permalink] 09 Jul 2014, 06:00
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