If 5 biscuits are distributed among 3 beggars, the chance that a particular beggar will get 2 biscuits is

Hint:To solve this question, we should write the sample space and the favourable cases. We should assume that every beggar gets at least one biscuit. Let us assume the biscuits given to the beggars as $\left( x,y,z \right)$. We should write all the cases by taking ${{x}_{1}}=1$. For example, we can write the cases as $\left( 1,1,3 \right),\left( 1,2,2 \right)\left( 1,3,1 \right)$. Likewise, we should write all the cases. For a beggar to get two biscuits, at least one value of ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ should be 2. In all the cases in sample space, we can get the favorable cases. The chance is given by the formula $\text{chance}=\dfrac{\text{favourable cases}}{\text{total cases}}$.

Complete step by step answer:

We are given that 5 biscuits are distributed to 3 beggars. We assume that every beggar gets at least one biscuit. Let us assume that $x,y,z$ are the biscuits given to the three beggars respectively.
Let the sample space which contains all the possible cases as S. Let the event E be when a particular beggar gets two biscuits. The condition that we should use is
$x+y+z=5$
Using this condition, we should write the sample space. The sample space can be written as
$S=\left\{ \left( 1,1,3 \right),\left( 1,3,1 \right),\left( 3,1,1 \right),\left( 2,2,1 \right),\left( 2,1,2 \right),\left( 1,2,2 \right) \right\}$
$n\left( S \right)=6$
The event E be when a particular beggar gets two biscuits. For a beggar to get two biscuits, at least one value of ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ should be 2. So, we should write the case from the sample space which has at least one of the values is 2.
$E=\left\{ \left( 2,2,1 \right),\left( 2,1,2 \right),\left( 1,2,2 \right) \right\}$
$n\left( E \right)=3$
We can write the required chance as
$\text{chance}=\dfrac{n\left( E \right)}{n\left( S \right)}=\dfrac{3}{6}=\dfrac{1}{2}$
$\therefore $The required chance is $\dfrac{1}{2}$

Note:
 We have relatively smaller values of the number of beggars and number of biscuits. If we have larger numbers, we can’t use this procedure. Let us assume that there are r beggars and n biscuits. We can write the condition as
$a+b+c+d....r\text{ terms}=n$. We have the formula for the number of non-zero solutions for the above equation as ${}^{n-1}{{C}_{r-1}}$. We have another formula for the number of solutions where zeros are allowed is ${}^{n+r-1}{{C}_{r-1}}$. Using these formulae, we can get the required values when the numbers are large.