Question

8 biscuits are distributed among 11 children find the probability that a particular child recieves 5 bisciuts

Answer 1

Thanks for A2A Shikhar Jaiswal.

8 biscuits can be distributed to 11 child in 11^8.

Since a specific child receives 5 biscuits, we can select a child in 11C1 ways.

Having given the 5 biscuits to a specific child, we are left with 3 biscuits.

These 3 biscuits are to be distributed among the remaining 10 beggars.

This can be done in 10^3 ways.

So summing it up, favourable cases are 11C1*(10^3)

total possible cases = 11^8

chance or probability = favourable cases/total number of cases

= [11C1*(10^3)]/[11^8]

Let us come to the answer provided by you. It looks like (I am just assuming), initially 5 biscuits were selected as understood by the term 11C5. The remaining 3 biscuits distributed among 10 people in 10^3 ways, as per the term in the numerator in the answer provided. The total number of cases are 11^8 in the denominator doesn't contradict our logical analysis.

But there is no mention about the biscuits being different. Had they been different or distinguishable from one biscuit from another, this term 11C5, would have been acceptable.

I hope this helps. :)