A purse contains n coins of unknown values. A coin drawn from it at random is found to be a rupee coin. Then the chance that it is the only rupee coin is?

The answer is given as $\frac{2}{n(n+1)}$

NOTE: Being "a rupee coin", means that the denomination of the coin is "one rupee". There are n coins of unknown denominations. I added this as the question might not have been very clear.

My approach for this question was to try to use the Bayes' Theorem, the hypothesis being the fact that the coin is the only rupee coin and the evidence (what is known to be true) is that the coin drawn is a rupee coin.

Let, $E$ be the event that the coin drawn is a rupee coin, $E_1$ be the event that it is the only rupee coin in the purse, $E_2$ be the event that it is not the only rupee coin in the purse.

Then, the probability that the coin is the only rupee coin given that it was drawn would be denoted by $P(E_1|E)$

Also, it is equally likely that the coin might or might not be the only rupee coin in the bag, giving us $P(E_1)=P(E_2)=\frac{1}{2}$

And, we get, $$P(E_1|E)=\frac{P(E_1)P(E|E_1)}{P(E_1)P(E|E_1)+P(E_2)P(E|E_2)}$$

But, I don't know and cant figure out what $P(E|E_1)$ and $P(E|E_2)$ would be here. Any help on that would be appreciated along with anyone if possible confirming the way that I've framed this problem so far is correct or not. And if this problem can be done with Bayes', how do I proceed?

  • $\begingroup$ "the chance that it is the only rupee coin is?" - I guess I don't understand. So, there could be dollars, cents, rupees, sums, etc. in the container? But you said unknown values. $\endgroup$
    – VIVID
    Jul 23, 2021 at 20:18
  • $\begingroup$ @VIVID So a rupee coin denotes "one rupee". There may be coins of 2 rupees, 3 rupees....x rupees, but it "being the only rupee coin" means that it has to be of one rupee. I understand how the language of the question can be confusing but I wrote the question as stated in the book, I will, however, add this into the question. $\endgroup$
    – Techie5879
    Jul 23, 2021 at 20:20
  • $\begingroup$ Ok, now it is clearer. Please, add this extra information into your post. $\endgroup$
    – VIVID
    Jul 23, 2021 at 20:21
  • 3
    $\begingroup$ I don't think one can answer this question without making some assumption about the distribution of the types of coins in the bag. For instance, you arbitrarily assumed $P(E_1)=P(E_2)=1/2$. $\endgroup$
    – angryavian
    Jul 23, 2021 at 20:28
  • 2
    $\begingroup$ That's not a correct assumption. $\endgroup$
    – Math Lover
    Jul 23, 2021 at 20:32

1 Answer 1


Let $X$ represent the number of one-rupee coins in the purse. Let's further assume that $X$ has a uniform distribution, i.e., $P(X = 1) = P(X = 2) = ... = P(X = n)$. So, all possible numbers of one-rupee coins are equally likely.

Note: This is an additional assumption that is not in the question, but as the comments point out, we have to have some sort of assumption on this distribution, and this assumption works.

Let $E$ be the observation (a one-rupee coin was removed from the purse). Then, from Bayes' Rule, we have

$$ P[X = 1 \ | \ E] = P[E \ | \ X = 1]\cdot \frac{P[X = 1]}{P[E]} . $$

From our uniform distribution assumption, we have $P[X = 1] = \frac{1}{n}$.

The conditional probability of observing a one-rupee coin is $P[E \ | \ X=x] = \frac{x}{n}$, so we have $P[E \ | \ X = 1] = \frac{1}{n}$.

The overall probability of observing a one-rupee coin is the sum over all possible cases, so we have:

$$ \begin{align} P[E] & = \sum_{x=1}^n {P[E \ | \ X=x]\cdot P[X=x]} \\ & = \sum_{x=1}^n {\frac{x}{n}\cdot\frac{1}{n}} \\ & = \frac{1}{n^2} \sum_{x=1}^n {x} \\ & = \frac{1}{n^2} \cdot \frac{n(n+1)}{2} \\ & = \frac{n+1}{2n}. \end{align} $$

Putting these three together, we have

$$\begin{align} P[X = 1 \ | \ E] &= \frac{1}{n}\cdot\frac{\frac{1}{n}}{\frac{n+1}{2n}} \\ & = \frac{1}{n}\cdot\frac{1}{n}\cdot\frac{2n}{n+1} \\ & = \frac{2}{n(n+1)}. \end{align}$$


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