Question

Two different families a and b are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families, so that no child gets more than one ticket. If the probability that all the tickets go to the children of family b is , then the number of children in each family is

Answer 1

Answer:

5

Step-by-step explanation:

Let 'x be the number of children in each family

i.e., Family 'a' has 'x' children and Family 'b' has 'x' children.

Given that 3 tickets to be distributed amongst the children such that no child gets more than 1 , which is equivalent to selecting a child and distributing 1 ticket to him.

Given that the probability that all the tickets go to the children of family 'b' is 1/12

[Please note 1/12 is missing in the question part.]

Hi

In total , there are 2x children, the number of ways of selecting 3 children

out of 2x children is given by 2xC3 ways.

But, given that all the children should be from family 'b'. Since there are 3

children in family 'b', we can select 3 children from family 'b' in xC3 ways.

Thus, the probability of all tickets going to children of family 'b' is given by

xC3/2xC3 = 1/12

=> x(x-1)(x-2)/2x(2x-1)(2x-2) = 1/12

=>(x-2)/4(2x-1) = 1/12

=> x-2  = (2x-1)/3

=> 3x - 6 = 2x - 1

=> x = 5.

Hence , there are 5 children in each family.

Hope, it helped !