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

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

Table pasted below to represent all 9 scenarios.
Attachments

table.PNG [ 263.31 KiB | Viewed 8932 times ]

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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
Nice question. I did it the following way:

Pairs for when z is odd
4,5 4,9 6,5 6,9 7,8
Each pair has probability of 1/3
So 1/3 * 1/3 * 1/3 ...5 times (5 pairs) = 5/9

Pairs for when z is even
4,8 6,8 6,5 6,9 7,8

Again each pair has probability of 1/3 ... 4 pairs = 4/9

5/9 - 4/9 = 1/9

Bunuel I tried to do this question through combination but wasn't able to. Any idea how to do that? TIA
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
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