The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. # 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.

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## Warm-up Questions

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

## Hints

• Hint 1
Suppose the first tin becomes empty, from which tin must the last biscuit be drawn from?
• Hint 2
Find 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 5
Does 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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