Q.(a) A bottle is full of dettol. One-third of its dettol is taken away and an equal amount of water is poured into the bottle to fill it again. This operation is repeated three times. Find the final ratio of dettol to water in the bottle.

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. …