Is my interpretation on the specific replacement in a solution is correct or not - Mixtures and Concentrations? Here is the question and solution -
A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process  was repeated further two times . How much milk is now contained by the container.
Solution -- 
Dilution -- concentration of 
milk is 36/40 - again this is two times more than - the net concentration of milk is 36/40 * 36/40 * 36/40.
Thus, the net amount of milk in 40 litres is 40 * 36^3/ 40^3.

My doubt is how they can assume that the concentration of milk at second time is 36/40. After the replacement, the milk concentration level is already 36/40. Now, we will be further diluting it. How is this possible to have milk concentration 36/40 each times? If I correlate this with the counting - permutation - if the events are dependent then the next event will be reduced from the 36/40 part. It cannot be independent events - not every time we will be taking out exactly 4 litres of milk. Next time, it is a solution and not a pure milk. How is this answer possible? If this is not the correct answer then please tell me the appropriate answer to this question. 
 A: Each dilution is achieved by replacing $4~\text{L}$ of the solution with $4~\text{L}$ of water.  
The first dilution reduces the concentration of milk to 
$$\frac{(40 - 4)~\text{L}}{40~\text{L}} = \frac{36}{40} = \frac{9}{10} = 0.9$$
Since 
$$\frac{9}{10} \cdot 40~\text{L} = 36~\text{L}$$
you now have $40~\text{L}$ of a solution that contains $36~\text{L}$ of milk. When you dilute it the second time, you pour off $4~\text{L}$ of this $90\%$ milk solution, thereby removing an additional
$$0.9(4~\text{L}) = 3.6~\text{L}$$ 
of milk, which is replaced by water, leaving you with $(40 - 4 - 3.6)~\text{L} = 32.4~\text{L}$ of milk.  The concentration of milk after two dilutions is  
$$\frac{32.4~\text{L}}{40~\text{L}} = \frac{0.9(36)}{40} = \frac{0.9[0.9(40)]}{40} = 0.9^2 = 0.81$$
In the third dilution, you replace $4~\text{L}$ of this $81\%$ milk solution with $4~\text{L}$ of water, thereby removing an additional 
$$0.81(4~\text{L}) = 3.24~\text{L}$$ 
of milk from the solution, leaving you with $(40 - 4 - 3.6 - 3.24)~\text{L} = 29.16~\text{L}$ of milk in the solution.  Thus, the concentration of milk after three dilutions is 
$$\frac{29.16~\text{L}}{40~\text{L}} = \frac{0.9(32.4)}{40} = \frac{0.9[0.9(36)]}{40} = \frac{0.9\{0.9[0.9(40)]\}}{40} = 0.9^3 = 0.729$$ 
Thus, the concentration of milk in the solution is reduced to $90\%$ after the first dilution, $81\%$ after the second dilution, and $72.9\%$ after the third dilution.  The reason the concentration is $9/10$ as much as it was previously after each dilution is that by replacing $1/10$ of the solution by water each time, you are replacing $1/10$ of the milk in the solution with water leaving you with $1 - 1/10 = 9/10$ as much milk in the solution as you had previously.
