Q.A container has 50 l of juice in it. 5 l of juice is taken out and is replaced by 5 l of water. This process is repeated 4 more times. What is the amount of juice in the container after final replacement?

Successive Replacement

Imagine you have a bag of 10 marbles — 4 red and 6 blue. You draw one marble, note its colour, and put it back. Then you draw another. The chance that both draws are red is $\frac{4}{10} \times \frac{4}{10} = 0.16$. That's straightforward because the bag is exactly the same for the second draw.

Now change the game: you draw a marble and do not put it back. The bag now has only 9 marbles. If the first was red, only 3 reds remain; if it was blue, all 4 reds are still there. The probability of the second event depends on what happened first. This is the core of successive replacement — or rather, its absence.

Successive replacement is the process of drawing items one after another from a finite set, returning each item to the set before the next draw. The key consequence: the composition of the set never changes. Every draw is identical and independent of every other draw.

The Precise Statement

Let a set contain $N$ items, of which $K$ are of a particular type (say "success"). You draw $n$ items one at a time, replacing each item before the next draw. Then:

• The probability of getting a success on any single draw is always $\frac{K}{N}$.
• The draws are independent — the outcome of one draw does not affect the next.
• The number of successes in $n$ draws follows a binomial distribution:

Why It Matters

Without replacement, probabilities shift after every draw — that's the hypergeometric distribution, and it's messier. With replacement, you get the clean, constant-probability binomial model. This is why successive replacement is the default assumption in many exam problems unless "without replacement" is explicitly stated. …