Question

If n biscuits are distributed among N different beggars ,what is the probability that a particular beggar gets exactly r biscuits ?
(A) \[\dfrac{{{}^n{C_r}{{\left( {N - 1} \right)}^r}}}{{{N^n}}}\]
(B) \[\dfrac{{{}^n{C_r}{{\left( {N - 1} \right)}^{n - r - 1}}}}{{{N^n}}}\]
(C) \[\dfrac{{{}^n{C_r}{{\left( {N - 1} \right)}^n}}}{{{N^n}}}\]
(D) \[\dfrac{{{}^n{C_r}{{\left( {N - 1} \right)}^{n - r}}}}{{{N^n}}}\]

Answer 1

Hint: To solve probability question we need total cases and favourable cases .Favourable case for this question is \[n\]biscuits are distributed among \[N\]different vectors and a particular beggar gets exactly \[r\]biscuit .Here to obtain total cases and favourable cases we will use concept of permutation and combination .
FORMULA USED :
Probability = \[\dfrac{{favourable\,cases}}{{Total\,cases}}\]
\[n\] things can be distributed among \[N\] people in total \[{N^n}\] ways .
We can choose \[r\] things from total \[n\] things in \[{}^n{C_r}\] ways .

Complete step by step solution:
Total no of ways we can distribute \[n\] biscuits in \[N\] beggars is =\[{N^n}\]
As the first biscuit can be given to \[N\] ways, the second biscuit again can be given to \[N\]ways and so on up to \[{n^{th}}\] biscuits .
This can be written mathematically =\[N \times N \times N \times N...........................n\,times\]
=\[{N^n}\]
Total no of biscuits are \[n\].
From these \[n\] biscuits we have to give \[r\] biscuits to a particular beggar .
We can choose \[r\] biscuits from total \[n\] biscuits in \[{}^n{C_r}\] ways .
Here the arrangements of the biscuits are not important so we use combinations .
From Total number of beggars one will get a particular number of biscuits. So Number of remaining beggars is one less than the total number of beggars .
So now number of remaining beggars are = \[N - 1\]
Rest units of the biscuits are =\[n - r\]
So now \[n - r\]biscuits have to be distributed among \[N - 1\] beggars .
Total no of ways we can distribute \[n - r\] biscuits in \[N - 1\] beggars is =\[{\left( {N - 1} \right)^{n - r}}\]
Required Probability = \[\dfrac{{favourable\,cases}}{{Total\,cases}}\]
Total cases =\[{N^n}\]
Favourable cases = \[{}^n{C_r}{\left( {N - 1} \right)^{n - r}}\]
So probability is = \[\dfrac{{{}^n{C_r}{{\left( {N - 1} \right)}^{n - r}}}}{{{N^n}}}\]
Option D is the correct answer .
So, the correct answer is “Option D”.

Note: Careful about permutation and combination . Depending upon the case we need to choose which one to apply. For Small number of beggars and biscuits students can write sample space and then calculate probability by counting total number of cases and favourable cases .