A glass is filled with milk. Two–thirds of its content is poured out and replaced with water. If this process of pouring out two–thirds the content and replacing with water is repeated three more times, then the final ratio of milk to water in the glass, is

The correct option is 1 : 80.

Let us understand the problem step-by-step. Let the total volume of the glass be represented by:
$V$
Initially, the glass is completely filled with milk. Therefore, the initial volume of milk is:
$V$
and the volume of water is:
$0$

In each cycle of the process, two-thirds of the glass's contents are poured out and replaced with water. The fraction poured out is:
$\frac{2}{3}$
This means that after each replacement, the remaining fraction of the liquid (and thus the remaining fraction of milk) is:
$1-\frac{2}{3}=\frac{1}{3}$
of the quantity present before that cycle.

The process is performed once initially, and then repeated three more times. This means the process is carried out a total of:
$4$
times.

Let us calculate the fraction of milk remaining after each of the operations:
1. After the 1st operation, milk remaining is:
$V\times \frac{1}{3}$
2. After the 2nd operation, milk remaining is:
$V\times {\left(\frac{1}{3}\right)}^2$
3. After the 3rd operation, milk remaining is:
$V\times {\left(\frac{1}{3}\right)}^3$
4. After the 4th operation, milk remaining is:
$V\times {\left(\frac{1}{3}\right)}^4$

Evaluating the final volume of milk:
$\text{Final Volume of Milk}=V\times \frac{1}{81}=\frac{V}{81}$

Since the glass is always filled back to its full capacity by adding water, the total volume of the mixture remains:
$V$
Therefore, the final volume of water in the glass is:
$\text{Final Volume of Water}=V-\frac{V}{81}=\frac{80V}{81}$

Now, we find the final ratio of milk to water in the glass:
$\text{Ratio of Milk to Water}=\frac{\text{Final Volume of Milk}}{\text{Final Volume of Water}}=\frac{V/81}{80V/81}=\frac{1}{80}$
Thus, the final ratio of milk to water is 1 : 80.