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

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

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

by BTGmoderatorLU » Tue Sep 11, 2018 2:40 pm

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Source: Manhattan Prep

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 OA is A.

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Source: Beat The GMAT — Problem Solving | Tue Sep 11, 2018 2:40 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 am; Location: Grand Central / New York; Thanked: 470 times; Followed by:34 members

by Jay@ManhattanReview » Tue Sep 11, 2018 9:56 pm

BTGmoderatorLU wrote:Source: Manhattan Prep

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 OA is A.

Since the ages are integers, their sum would be either even or odd.

We see that there are three boys and three girls, thus, there are 3*3 = 9 possible sums; out of which few are even and few are odd.

Let's first find out the sum = Odd.

Sum would be odd if an even adds to odd OR and odd adds to even.

1. Say, we pick one boy with even age, thus, we must pick a girl with odd age. There are two boys with even age (4 and 6) and there are two girls with odd age (5 and 9).

> Number of ways to select one out of two boys = 2C1 = 2
> Number of ways to select one out of two girls = 2C1 = 2

Total number of ways = 2*2 = 4

2. Say, we pick one boy with odd age, thus, we must pick a girl with even age. There is only one boy with even age (7) and there is only one girl with odd age (8).

> Number of ways to select one out of one boy = 1
> Number of ways to select one out of one girl = 1

Total number of ways = 1*1 = 2

Total number of ways to get odd sum = 4 + 1 = 5

=> The probability that z is odd = 5/9

=> The probability that z is even = 1 - 5/9 = 4/9

Difference = |4/9 - 5/9| = 1/9

The correct answer: A

Hope this helps!

-Jay
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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Sep 12, 2018 3:19 am

BTGmoderatorLU wrote:Source: Manhattan Prep

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

Since only a few cases are possible, we can list them out:
4+6+5+8 = 23
4+6+5+9 = 24
4+6+8+9 = 27
4+7+5+8 = 24
4+7+5+9 = 25
4+7+8+9 = 28
6+7+5+8 = 26
6+7+5+9 = 27
6+7+8+9 = 30

Of the 9 cases above, the four in red are ODD, implying that the remaining five cases are EVEN.
Since 5/9 of the cases are even, and 4/9 are odd, we get:
P(z is even) - P(z is odd) = 5/9 - 4/9 = 1/9.

The correct answer is A.

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by Scott@TargetTestPrep » Wed Sep 12, 2018 5:34 pm

BTGmoderatorLU wrote:Source: Manhattan Prep

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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