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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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New post Updated on: 09 Dec 2012, 09:09
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A
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E

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Question Stats:

70% (02:52) correct 30% (02:48) wrong based on 261 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?

(A) 1/9
(B) 1/6
(C) 2/9
(D) 1/4
(E) 1/2

Originally posted by mun23 on 09 Dec 2012, 09:05.
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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New post 09 Dec 2012, 09:27
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2
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]

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New post 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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New post 23 Apr 2015, 13:32
1
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

Answer=5/9 - 4/9 = 1/9

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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New post 13 Jul 2015, 11:06
1
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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New post 13 Mar 2017, 00:48
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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
table.PNG [ 263.31 KiB | Viewed 1755 times ]


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

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New post 12 Sep 2018, 17:36
1
mun23 wrote:
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




The sum of the two selected boys can be either 4 + 6 = 10, 4 + 7 = 11 or 6 + 7 = 13. Thus, there is a 1/3 probability that the sum of the ages of the two boys will be even and 2/3 probability that the sum of the ages of the two boys will be odd.

Similarly, the sum of the two selected girls can be either 5 + 8 = 13, 5 + 9 = 14 or 8 + 9 = 17. Thus, there is a 1/3 probability that the sum of the ages of the two girls will be even and 2/3 probability that the sum of the ages of the two girls will be odd.

Now, let’s first find the probability that z is even. Since z is the sum of ages of the selected boys and girls, z can be even if the sum of both the selected boys and selected girls ages are even or if sum of both the selected boys and selected girls ages are odd. The probability that both sums are even is 1/3 x 1/3 = 1/9 and the probability that both sums are odd is 2/3 x 2/3 = 4/9. Thus, there is a 1/9 + 4/9 = 5/9 probability that z is even.

Similarly, let’s find the probability that z is odd. z can only be odd of one of the sums is even and the other is odd. Probability that the boys sum is even and girls sum is odd is 1/3 x 2/3 = 2/9. Probability that the boys sum is odd and the girls sum is even is 2/3 x 1/3 = 2/9. Thus, there is a 2/9 + 2/9 = 4/9 probability that z is odd.

Finally, the difference between the probabilities that z is even and z is odd is 5/9 - 4/9 = 1/9.

Answer: A
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Re: Three boys are ages 4, 6 and 7 respectively. Three girls are &nbs [#permalink] 12 Sep 2018, 17:36
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