Question

If 10 distinct objects are distributed among 3 persons find the chance of a particular person having more than 5 of them.

Answer 1

distributing 10 distinct object among 3 person can be done in 3^10 ways(think from the point of view of an object each got 3 destinations,then apply product rule)
in the case where a particular person having more than 5
he can have 6 7 8 9 and 10
if he has 6
he can be chosen in 3 ways
first we have to select the 6 objects out of the 10
this can be done in 10c6 ways
and then distribute other 4objects to the other 2 persons.this can be done in 2^4 ways(Again think from the side of objects each has got 2 destinations.)
3*10c6*2^4
similarly if he has 7
then this can be
3*10c7*2^3
if it is 8
3*10c8*2^2
if it is 9
3*10c9*2
if it is 10
then
3*10c10*2^0
so no of favourable cases =

3*10c10*2^0+3*10c9*2^1+3*10c8*2^2+3*10c7*2^3+3*10c6*2^4
total no of cases=3^10
probability =no of favourable cases /no of total case=.2297