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Vagrant
Dec6-06, 09:36 AM
Hi, I'd appriciate 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

quasar987
Dec6-06, 11:18 AM
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.

Vagrant
Dec7-06, 06:30 AM
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.

Muzza
Dec7-06, 08:43 AM
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?

Vagrant
Dec8-06, 12:43 PM
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.

ssd
Dec20-06, 11:30 AM
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.