Question

K balls are distributed at random and independently of one another among N cells which lie in a straight line $(N>K).$ Find the probability that they will occupy k adjacent cells.

• A. $\frac{(N-1)}{N^k}$
• B. $\frac{(N-K-1)}{N^k}$
• C. $\frac{(N-K)}{N^k}$
• D. $\frac{(N-K+1)}{N^k}$

Answer 1

Answer: (D)

$\frac{(N-K+1)}{N^k}$

$n=$ the total number of ways of distributing k balls over N cells $=N^k$
Now we find the favorable no .of ways .K adjacent cells out of
N can be chosen in N-K+1ways.
For if we denote the N .cells by $C_1,C_2,C_3,\cdots ,C_N$
then the following groupings of K consecutive cells are possible.
$\begin{array}{cc}{C}_1{C}_2\cdots \cdots & C_k\\ C_2{C}_3\cdots \cdots & C_{k+1}\\ C_3{C}_4\cdots \cdots & C_{k+2}\end{array}$
$C_{N-K},C_{N-K+1},....C_{N-1},$ $C_{N-K+1},C_{N-K+2},....C_N.$
Now k balls can be distributed over each of these groups of k consecutive cells in $K!$ ways.
Hence $m=$ the favourable no .of ways $=(N-K+1)K!$
$\therefore$ The required probability $=\frac{(N-K+1)}{N^k}$