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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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Kudos
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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The provided text is a discussion thread from a GMAT forum, where users are discussing a probability problem involving the ages of three boys and three girls. The problem asks for the difference in probabilities between the sum of the ages of two randomly chosen boys and two randomly chosen girls being even or odd.

### Content Structure:
- The text starts with metadata about the forum post, including the date and time.
- It then presents the problem statement and the ages of the boys and girls.
- Following the problem statement, there are multiple user posts with different approaches to solving the problem.

### Analysis:
single_question_with_answer

- **no_question**: the text clearly contains a mathematical question.
- **single_question**: there is one main question being discussed.
- **single_question_with_answer**: multiple users provide different solutions and answers to the question.
- **multiple_questions**: the text does not contain multiple distinct questions.
- **multiple_questions_with_answers**: the text does not contain multiple distinct questions with answers.

the text contains a single mathematical question, and multiple users provide different solutions and answers to this question. therefore, the appropriate type is **single_question_with_answer**.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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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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Kudos
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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Kudos
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]
2
Bookmarks
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]
1
Kudos
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]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
Moderator:
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94441 posts
user
The provided text is a discussion thread from a GMAT forum, where users are discussing a probability problem involving the ages of three boys and three girls. The problem asks for the difference in probabilities between the sum of the ages of two randomly chosen boys and two randomly chosen girls being even or odd.

### Content Structure:
- The text starts with metadata about the forum post, including the date and time.
- It then presents the problem statement and the ages of the boys and girls.
- Following the problem statement, there are multiple user posts with different approaches to solving the problem.

### Analysis:
TYPE: SINGLE_QUESTION_WITH_ANSWER

- **NO_QUESTION**: The text clearly contains a mathematical question.
- **SINGLE_QUESTION**: There is one main question being discussed.
- **SINGLE_QUESTION_WITH_ANSWER**: Multiple users provide different solutions and answers to the question.
- **MULTIPLE_QUESTIONS**: The text does not contain multiple distinct questions.
- **MULTIPLE_QUESTIONS_WITH_ANSWERS**: The text does not contain multiple distinct questions with answers.

The text contains a single mathematical question, and multiple users provide different solutions and answers to this question. Therefore, the appropriate type is **SINGLE_QUESTION_WITH_ANSWER**.