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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]
  2
  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]
  1
  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]
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  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!
  
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  Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
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explanation: The provided text is a GMAT forum discussion about a probability problem. It includes a question and multiple user responses with different approaches to solving the problem. The core question is:
  
  **Question:**
  Three boys are ages 4, 6, and 7 respectively. Three girls are ages 5, 8, and 9 respectively. If two boys and two girls are chosen at random, what is the probability that the sum of their ages is even? What is the difference between the probability that the sum of their ages is even and the probability that the sum of their ages is odd?
  
  **User Responses:**
  - Users provide various methods to solve the problem, including combinatorial approaches and probability calculations.
  - The final answer is that the difference between the probability that the sum of their ages is even and the probability that the sum of their ages is odd is \( \frac{1}{9} \).
  
  **
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Last visit was: 21 Jul 2024, 20:43 It is currently 21 Jul 2024, 20:43
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  # Three boys are ages 4, 6 and 7 respectively. Three girls are
  
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  Manager
  Joined: 06 Dec 2012
  Status:struggling with GMAT
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  Concentration: Accounting
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  Manager
  Joined: 31 May 2012
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  Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
  2
  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
  Senior Manager
  Joined: 02 Dec 2014
  Posts: 304
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  GMAT 1: 640 Q44 V33
  WE:Sales (Telecommunications)
  Re: Three boys are ages 4, 6 and 7 respectively. Three girls are [#permalink]
  1
  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)?
  Manager
  Joined: 04 May 2015
  Posts: 64
  Own Kudos [?]: 30 [2]
  Given Kudos: 58
  Concentration: Strategy, Operations
  WE:Operations (Military & Defense)
  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.
  Director
  Joined: 26 Oct 2016
  Posts: 506
  Own Kudos [?]: 3411 [1]
  Given Kudos: 877
  Location: United States
  Schools: HBS '19
  GMAT 1: 770 Q51 V44
  GPA: 4
  WE:Education (Education)
  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 ]
  
  Manager
  Joined: 20 Mar 2019
  Posts: 146
  Own Kudos [?]: 15 [0]
  Given Kudos: 282
  Location: India
  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
  Non-Human User
  Joined: 09 Sep 2013
  Posts: 34040
  Own Kudos [?]: 853 [0]
  Given Kudos: 0
  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:
  Math Expert
  94441 posts
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a GMAT forum discussion about a probability problem. It includes a question and multiple user responses with different approaches to solving the problem. The core question is:
  
  **Question:**
  Three boys are ages 4, 6, and 7 respectively. Three girls are ages 5, 8, and 9 respectively. If two boys and two girls are chosen at random, what is the probability that the sum of their ages is even? What is the difference between the probability that the sum of their ages is even and the probability that the sum of their ages is odd?
  
  **User Responses:**
  - Users provide various methods to solve the problem, including combinatorial approaches and probability calculations.
  - The final answer is that the difference between the probability that the sum of their ages is even and the probability that the sum of their ages is odd is \( \frac{1}{9} \).
  
  **Label:**
  question-and-answer