$n$ 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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- 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?
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Suppose the first tin becomes empty, from which tin must the last biscuit be drawn from?
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Find the probability that there is 1 biscuit left in the first tin and
$k$biscuits left in the second tin. -
… 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. -
… and then the number of possible orderings which lead to that scenario.
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Does it matter that which tin becomes empty and which tin contains
$k$biscuits?
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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
$2{2n - k - 1 \choose n - 1}(\frac{1}{2})^{2n-k}.$