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Question: If n biscuits are distributed among N boys at random. The probability that a particular boy gets r (...

If n biscuits are distributed among N boys at random. The probability that a particular boy gets r (< n) biscuits is:
(a) nCr(1N)r(N−1N)n−r^{n}{{C}_{r}}{{\left( \dfrac{1}{N} \right)}^{r}}{{\left( \dfrac{N-1}{N} \right)}^{n-r}}
(b) nCr(1N)r^{n}{{C}_{r}}{{\left( \dfrac{1}{N} \right)}^{r}}
(c) nCr^{n}{{C}_{r}}
(d) rn\dfrac{r}{n}

Explanation

Solution

Hint: To solve this question, we should know the concept of the combination, that is the number of ways of combining things irrespective of their orders and it can be expressed as nCr^{n}{{C}_{r}} in mathematical terms. Also, we should have some knowledge of the probability that is the ratio of the favorable number of outcomes of events to the total number of outcomes. Mathematically, we can write it as,
Probability = Favorable OutcomesTotal Outcomes\text{Probability = }\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}

Complete step-by-step answer:
In this question, we have to find the probability that a particular boy gets r (< n) biscuits. In this question, we have been given that n biscuits are distributed among N boys. So, we can say that each biscuit can be given to any one of the N boys. So, the total number of ways of distributing n biscuits among N boys is N×N×N×.....N\times N\times N\times .....n times.
=Nn ways={{N}^{n}}\text{ ways}
Or, we can say the total number of outcomes of distributing n biscuits among N boys is Nn{{N}^{n}} ways. Now, we have been given that a particular boy gets r biscuits. So, by the method of combination, we can say there are nCr^{n}{{C}_{r}} ways to choose r biscuits out of n biscuits. Now, we are left with (n – r) biscuits and we have to distribute them among (N – 1) boys. So, the number of ways of distributing them can be
(N−1)×(N−1)×.....(n−r) times = (N−1)n−r\left( N-1 \right)\times \left( N-1 \right)\times .....\left( n-r \right)\text{ times = }{{\left( N-1 \right)}^{n-r}}
Or we can say the favorable outcomes of distributing n biscuits among N boys such that one boy gets r ( < n) biscuits = nCr(N−1)n−r^{n}{{C}_{r}}{{\left( N-1 \right)}^{n-r}}. Now, we have to find the probability of a particular boy getting r biscuits. And, we know that the probability of an event is Favorable OutcomesTotal Outcomes\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}.
So, the probability of a particular boy getting r biscuits is P,
P=nCr(N−1)n−rNnP=\dfrac{^{n}{{C}_{r}}{{\left( N-1 \right)}^{n-r}}}{{{N}^{n}}}
P= nCr(N−1)n−rNn+r−rP={{\text{ }}^{n}}{{C}_{r}}\dfrac{{{\left( N-1 \right)}^{n-r}}}{{{N}^{n+r-r}}}
P= nCr(N−1)n−rNrNn−rP={{\text{ }}^{n}}{{C}_{r}}\dfrac{{{\left( N-1 \right)}^{n-r}}}{{{N}^{r}}{{N}^{n-r}}}
P= nCr(1N)r(N−1N)n−rP={{\text{ }}^{n}}{{C}_{r}}{{\left( \dfrac{1}{N} \right)}^{r}}{{\left( \dfrac{N-1}{N} \right)}^{n-r}}
Hence, we can see that option (a) is the only right answer.

Note: The possible mistake one can make is in choosing the total number of outcomes and favorable outcomes. Also, the possible mistake one can do is by writing Nn{{N}^{n}} as nN{{n}^{N}} or (N−1)n−r{{\left( N-1 \right)}^{n-r}} as (n−r)N−1{{\left( n-r \right)}^{N-1}}. Also, one might choose option (b) as the correct answer but that is wrong because in that case (n – r) biscuits are distributed among N boys.