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

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Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]

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09 Dec 2012, 09:05
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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]

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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

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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]

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01 Jan 2013, 14:19
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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]

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09 Jul 2014, 06:00
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]

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23 Apr 2015, 13:32
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

I think 13 should be here not 12. And i don't understand second bolded fragment. May be this should be (26,30,27)?
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]

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13 Jul 2015, 11:06
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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.
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]

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19 Sep 2016, 08:16
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are   [#permalink] 19 Sep 2016, 08:16
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