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2007 Paper 1 Q12
D: 1500.0 B: 1484.0
probability conditional probability without replacement law of total probability symmetry combinatorics transfer problems

  1. A bag contains \(N\) sweets (where \(N \ge 2\)), of which \(a\) are red. Two sweets are drawn from the bag without replacement. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.
  2. There are two bags, each containing \(N\) sweets (where \(N \ge 2\)). The first bag contains \(a\) red sweets, and the second bag contains \(b\) red sweets. There is also a biased coin, showing Heads with probability \(p\) and Tails with probability \(q\), where \(p+q = 1\). The coin is tossed. If it shows Heads then a sweet is chosen from the first bag and transferred to the second bag; if it shows Tails then a sweet is chosen from the second bag and transferred to the first bag. The coin is then tossed a second time: if it shows Heads then a sweet is chosen from the first bag, and if it shows Tails then a sweet is chosen from the second bag. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.

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2007 Paper 1 Q13
D: 1500.0 B: 1469.5
probability conditional probability combinatorics without replacement counting discrete probability

A bag contains eleven small discs, which are identical except that six of the discs are blank and five of the discs are numbered, using the numbers 1, 2, 3, 4 and 5. The bag is shaken, and four discs are taken one at a time without replacement. Calculate the probability that:

  1. all four discs taken are numbered;
  2. all four discs taken are numbered, given that the disc numbered ``3'' is taken first;
  3. exactly two numbered discs are taken, given that the disc numbered ``3'' is taken first;
  4. exactly two numbered discs are taken, given that the disc numbered ``3'' is taken;
  5. exactly two numbered discs are taken, given that a numbered disc is taken first;
  6. exactly two numbered discs are taken, given that a numbered disc is taken.

Solution: There are many ways to do the counting in each question, possibly the clearest way is to always consider the order in which discs are taken, although all methods should work equally well. For some examples Bayes rule also offers a fast solution.

  1. There are we are choose \(4\) objects in order from \(5\) (ie \({^5\P_4}\)) to obtain valid draws, this is out of a total of picking \(4\) objects from \(11\) (\({^{11}\P_4}\)). Ie the probability is: \(\displaystyle \frac{{^5\P_4}}{{^{11}\P_4}} = \frac{5! \cdot 7!}{11!} = \frac{1}{66}\) Alternatively, there are \(\binom{5}{4}\) ways to choose four numbered discs, out of \(\binom{11}{4}\) ways to choose four discs. ie \(\displaystyle \binom{5}{4} \Big / \binom{11}{4} = \frac{5 \cdot 4! \cdot 7!}{11!} = \frac{5 \cdot 4 \cdot 3 \cdot 2}{11 \cdot 10 \cdot 9 \cdot 8} = \frac1{66}\)
  2. \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{^4\P_3 \big / {^{11}\P_4}}{1/11} \\ &= 11\cdot \frac{4!}{1!} \Bigg / \frac{11!}{7!} \\ &= \frac{4! \cdot 7! \cdot 11}{11!} \\ &= \frac{4\cdot 3 \cdot 2}{10 \cdot 9 \cdot 8} \\ &= \frac{1}{30} \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{3} \Bigg / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac{1}{30} \end{align*} Where we are calculating this as "choose one number", then "choose 3 more", which can happen ending up with 3, number, number, number in \(\binom{4}{3}\) ways, and there are \(11 \cdot \binom{10}{3}\) was overall. Another alternative using Bayes rule: \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \mathbb{P}( \text{first disc is 3} | \text{all four discs are numbered}) \frac{ \mathbb{P}( \text{all four discs are numbered} }{ \mathbb{P}( \text{first disc is 3} )} \\ &= \frac{\frac{1}{5} \cdot \frac{1}{66}}{\frac{1}{11}} \\ &= \frac1{30} \end{align*}
  3. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{3 \cdot {^4\P_1} \cdot {^{6}\P_2} \big / {^{11}\P_4}}{\frac1{11} } \\ &= \frac12 \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac12 \end{align*}
  4. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and 3 taken})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / \binom{11}{4}}{\frac{4}{11}} \\ &= \frac{\frac{2}{11}}{\frac4{11}} \\ &= \frac12 \end{align*} Using Bayes rule: \(\mathbb{P}( \text{3 taken}) = \frac{1}{11} + \frac{10}{11}\frac{1}{10} + \frac{10}{11}\frac{9}{10}\frac{1}{9} + \frac{10}{11}\frac{9}{10}\frac89\frac18 = \frac{4}{11}\) \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{3 taken | exactly two discs are numbered})\mathbb{P}(\text{exactly two discs are numbered})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\frac{4}{10} \cdot \binom{5}{2} \binom{6}{2} \Big / \binom{11}{4}}{4/11} \\ &= \frac{\frac4{10}{5 / 11}}{4/11} \\ &= \frac{1}{2} \end{align*}
  5. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc first}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc first})}{\mathbb{P}( \text{numbered disc first})} \\ &= \frac{3 \cdot {^5\P_1}\cdot{^4\P_1}\cdot{^6\P_2} \Big / {^{11}\P_4}}{\frac{5}{11}} \\ &= \frac{1}{2} \end{align*}
  6. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc taken})}{\mathbb{P}(\text{numbered disc taken})} \\ &= \frac{\mathbb{P}(\text{exactly two discs are numbered})}{1 - \mathbb{P}(\text{not numbered discs taken})} \\ &= \frac{\binom{5}{2}\binom{6}{2} \Big / \binom{11}{4}}{1 - \binom{6}{4} \Big / \binom{11}{4}} \\ &= \frac{\frac{5}{11}}{\frac{21}{22}} \\ &= \frac{10}{21} \neq \frac12 \end{align*}

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2004 Paper 2 Q13
D: 1600.0 B: 1500.0
probability conditional discrete probability distributions expected value optimisation without replacement hypergeometric sequential decision

A bag contains \(b\) balls, \(r\) of them red and the rest white. In a game the player must remove balls one at a time from the bag (without replacement). She may remove as many balls as she wishes, but if she removes any red ball, she loses and gets no reward at all. If she does not remove a red ball, she is rewarded with \pounds 1 for each white ball she has removed. If she removes \(n\) white balls on her first \(n\) draws, calculate her expected gain on the next draw and show that %her expected total reward would be the same as before it is zero if \(\ds n = {b-r \over r+1}\,\). Hence, or otherwise, show that she will maximise her expected total reward if she aims to remove \(n\) balls, where \[ n = \mbox{ the integer part of } \ds {b + 1 \over r + 1}\;. \] With this value of \(n\), show that in the case \(r=1\) and \(b\) even, her expected total reward is \(\pounds {1 \over 4}b\,\), and find her expected total reward in the case \(r=1\) and \(b\) odd.

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1991 Paper 3 Q15
D: 1700.0 B: 1485.9
probability combinatorics cards without replacement hypergeometric counting arguments pairs tree diagram

A pack of \(2n\) (where \(n\geqslant4\)) cards consists of two each of \(n\) different sorts. If four cards are drawn from the pack without replacement show that the probability that no pairs of identical cards have been drawn is \[ \frac{4(n-2)(n-3)}{(2n-1)(2n-3)}. \] Find the probability that exactly one pair of identical cards is included in the four. If \(k\) cards are drawn without replacement and \(2 < k < 2n,\) find an expression for the probability that there are exactly \(r\) pairs of identical cards included when \(r < \frac{1}{2}k.\) For even values of \(k\) show that the probability that the drawn cards consist of \(\frac{1}{2}k\) pairs is \[ \frac{1\times3\times5\times\cdots\times(k-1)}{(2n-1)(2n-3)\cdots(2n-k+1)}. \]

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