# Practice Paper 4 Question 10

There are two tins each containing \(n\) biscuits. A tin is chosen at random and a biscuit is removed. This is repeated until one tin becomes empty. What is the probability that there are \(k\) biscuits left in one tin when the other becomes empty? Explain the steps of your solution.

## Related topics

## Warm-up Questions

- A coin is flipped 5 times, displaying either heads or tails. What is the probability that you obtain heads exactly 3 times?
- You roll two dice. What is the probability that their sum is less than 6?

## Hints

- Hint 1Suppose the first tin becomes empty, from which tin must the last biscuit be drawn from?
- Hint 2Find the probability that there is 1 biscuit left in the first tin and \(k\) biscuits left in the second tin.
- Hint 3... by first finding the probabilities that \(n-1\) biscuits were drawn from the first tin and \(n-k\) biscuits were drawn from the second tin.
- Hint 4... and then the number of possible orderings which lead to that scenario.
- Hint 5Does it matter that which tin becomes empty and which tin contains \(k\) biscuits?

## Solution

It doesn't matter which tin becomes empty and which tin contains \(k\) biscuits due to symmetry.

Suppose it's the first one that becomes empty: the last biscuit must be drawn from that tin, which has probability \(\frac{1}{2}.\) Therefore, prior to that, there must have been \(n-1\) biscuits drawn from the first tin, which has probability \(\frac{1}{2^{n-1}},\) and \(n-k\) biscuits drawn from the other tin, giving the probability \(\frac{1}{2^{n-k}}.\) There are \({ (n - 1) + (n - k) \choose n - 1}\) possible orderings of those choices. This gives a probability of \({2n - k - 1 \choose n - 1}\big(\frac{1}{2}\big)^{2n-k}.\)

We get the same probability if it's the second tin that becomes empty. Those two events are independent, so the probabilities can be added. The final result is \(2{2n - k - 1 \choose n - 1}(\frac{1}{2})^{2n-k}.\)

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