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

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09 Dec 2012, 10:05

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67% (03:06) correct
33% (02:13) wrong based on 190 sessions

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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?

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

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09 Jul 2014, 07:00

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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, 12:06

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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, 09:16

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

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13 Mar 2017, 01:48

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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 [ 263.31 KiB | Viewed 182 times ]

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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are
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13 Mar 2017, 01:48

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