Question

A purse contains 100 coins of unknown value, a coin drawn at random is found to be a rupee. The chance that it is the only rupee in the purse is (assume all numbers of rupee coins in the purse to be equally likely.)
(a) ${\displaystyle \frac{1}{5050}}$
(b) ${\displaystyle \frac{2}{5151}}$
(c) ${\displaystyle \frac{1}{4950}}$
(d) ${\displaystyle \frac{2}{4950}}$

Answer 1

Hint: To solve this question we will first calculate the total number of cases that a number of one rupee coins in the bag can be 1, 2, 3, …, upto 100. So for each case the chance of drawing will be ${\displaystyle \frac{x}{n}}$ where x is the number of one rupee coin and n is total number of coins i.e. 100 so we will add all the cases using the fact that sum of n natural number is given by ${\displaystyle \frac{n(n+1)}{2}}$ . We will get the total chances of drawing one rupee coin after that we will divide the ${\displaystyle \frac{1}{100}}$ with the total chances obtained as we are given that it is the only rupee in the purse.

Complete step-by-step answer:
We are given that a purse contains 100 coins and when a coin is drawn out of it randomly it is found to be a one rupee coin so we have to find the chances that it is the only one rupee coin out of 100 coins in the purse.
So first of all we will find the total number of chances i.e. all the chance when there are
1, 2, 3, …, 100 one rupee coins in the purse.
So when there is only 1 one rupee coin chances to draw it out are = ${\displaystyle \frac{1}{100}}$
when there is only 2 one rupee coin chances to draw it out are = ${\displaystyle \frac{2}{100}}$
…..
when there is only 100 one rupee coin chances to draw it out are = ${\displaystyle \frac{100}{100}}$
so the total chances are = ${\displaystyle \frac{1}{100}}+{\displaystyle \frac{2}{100}}+{\displaystyle \frac{3}{100}}+....+{\displaystyle \frac{100}{100}}$
$={\displaystyle \frac{1+2+3+......+100}{100}}$
So to find the sum of first n natural numbers we know formula is ${\displaystyle \frac{n(n+1)}{2}}$, so we get
$\begin{array}{rl}& ={\displaystyle \frac{{\displaystyle \frac{100\left(101\right)}{2}}}{100}}\\ & ={\displaystyle \frac{101}{2}}\end{array}$
So we get the total number of cases as ${\displaystyle \frac{101}{2}}$
Now we have to find that it is the only coin so that can be done in ${\displaystyle \frac{1}{100}}$ ways,
Now the chance that it is the only coin = ${\displaystyle \frac{favourablecase}{Totalnumberofcases}}$
=${\displaystyle \frac{{\displaystyle \frac{1}{100}}}{{\displaystyle \frac{101}{2}}}}$
=${\displaystyle \frac{2}{100\times 101}}$
= ${\displaystyle \frac{2}{10100}}$
= ${\displaystyle \frac{1}{5050}}$

So, the correct answer is “Option (a)”.

Note: To solve this problem keep in mind that we have to count the total number of chances first i.e. the cases when there are a random number of one rupees coins in the purse. Some students might just find out the answer as ${\displaystyle \frac{1}{100}}$ but it is not what is asked in the question so be careful while solving this and read the question carefully.