How many children do Person X and Y have?

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Person X and Y each have 4 children, as determined through combinatorial probability. The problem involves distributing 3 movie tickets among 2N children, where N represents the number of children each parent has. The probability that 2 tickets go to the children of one parent and 1 ticket to the other is calculated to be 6/7, leading to the conclusion that N equals 4. The solution utilizes combinatorial expressions rather than Bayes' theorem.

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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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