How many children do Person X and Y have?

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Discussion Overview

The discussion revolves around a probability problem involving two individuals, Person X and Y, who have an equal number of children. The problem presents a scenario with movie tickets and asks for the number of children based on a given probability of ticket distribution.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant requests help with a problem involving the distribution of movie tickets among children of Person X and Y, noting that the probability of a specific distribution is 6/7.
  • Another participant suggests using combinatorial expressions to relate the number of children (N) to the given probability.
  • A different participant expresses confusion about the connection between the data provided and the number of children, indicating that both individuals must have at least two children.
  • One participant proposes starting with the total number of ways to distribute the tickets among the children of both individuals.
  • Another participant claims to have used Bayes' theorem to arrive at a conclusion that the number of children is 4, asserting this is correct.
  • A later reply challenges the necessity of Bayes' theorem, providing an alternative combinatorial approach that also leads to the conclusion of 4 children.

Areas of Agreement / Disagreement

There is no consensus on the method to solve the problem, with participants presenting different approaches and interpretations. Some agree on the conclusion of 4 children, while the necessity of specific methods remains contested.

Contextual Notes

The discussion includes varying assumptions about the distribution of tickets and the mathematical methods employed, with some participants questioning the relevance of certain approaches.

Vagrant
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Hi, I'd appreciate help with this problem;

Person X and Y have equal number of kids. There are 3 movie tickets. The probability that 2 tickets go to kids of one and 1 ticket goes to the kids of other is 6/7. How many kids do X and Y have?

Thanx
 
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Supposing each way of distributing the tickets is of equal probability, you can find and expression using combinatorics for the desired probability in terms of N (the number of kids). Then set the probability equal to 6/7 and solve for N.
 
sorry, but i don't see any connection between the data given and no. of kids, except that both have 2 or more kids. Could u please be more definite.
Thanx for ur help.
 
Let X have N children, and Y have N children. For starters, how many ways are there to distribute the 3 tickets among the 2N children?
 
Well, i used Baye's theorem n I'm getting the no. of children to be 4, which is correct, thank You for your help.
 
shramana said:
Well, i used Baye's theorem n I'm getting the no. of children to be 4, which is correct, thank You for your help.

Baye's theorem is not reqiured:
2(nC2)(nC1)/(2nC3)=6/7 . Solving we get, n=4.
 

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