Question

Two boys $A$ and $B$ find the jumble of n ropes lying on the floor. Each takes hold of one loose end randomly. If the probability that they are both holding the same rope is $\frac{1}{101}$, then the number of ropes is equal to

Answer 1

Let's find the number of ropes.

As we know that a rope has two ends.

Hence, for $n$ ropes, the total number of ends $=2n$.

So, remaining ends $=2n-1$.

Hence, selecting one end from $\begin{array}{rcl}2n-1& =& \frac{1}{\left(2n-{}^{1}C_1\right)}\\ & =& \frac{1}{\left(2n-1\right)}\\ \mathrm{So},2\mathrm{n}-1& =& 101\\ 2\mathrm{n}& =& 102\\ \mathrm{n}& =& \frac{102}{2}=51\end{array}$

Therefore, the number of ropes is equal to $51$.