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2009 Paper 1 Q13
D: 1500.0 B: 1504.1
probability discrete probability distributions expectation combinatorics arrangements counting arguments permutations

I seat \(n\) boys and \(3\) girls in a line at random, so that each order of the \(n+3\) children is as likely to occur as any other. Let \(K\) be the maximum number of consecutive girls in the line so, for example, \(K=1\) if there is at least one boy between each pair of girls.

  1. Find \(\P(K=3)\).
  2. Show that \[\P(K=1)= \frac{n(n-1)}{(n+2)(n+3)}\,. \]
  3. Find \(\E(K)\).

Solution:

  1. If all the girls are say together there are \(n+1\) ways to place the block of 3 girls. There are \(\binom{n+3}{3}\) ways to choose where to place the girls in total, therefore: \begin{align*} && \mathbb{P}(K =3) &= \frac{n+1}{\binom{n+3}3} \\ &&&= \frac{6(n+1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{6}{(n+3)(n+2)} \end{align*}
  2. If \(K= 1\) then all of the girls are separated. We can place three girls and two boys separating them, then we are allocating \(N-2\) boys to \(4\) gaps, ie \(\binom{N-2+3}{3} = \binom{N+1}{3}\). \begin{align*} && \mathbb{P}(K=3) &= \frac{\binom{n+1}{3}}{\binom{n+3}{3}} \\ &&&= \frac{(n+1)n(n-1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{n(n-1)}{(n+3)(n+2)} \end{align*}
  3. \(\,\) \begin{align*} \mathbb{E}(K) &= \sum_{k=1}^3 k \mathbb{P}(K=k) \\ &= \frac{6}{(n+3)(n+2)} + 2 \left (1 - \frac{6}{(n+3)(n+2)} - \frac{n(n-1)}{(n+3)(n+2)} \right) + 3\frac{n(n-1)}{(n+3)(n+2)} \\ &= 2+\frac{6-12+n(n-1)}{(n+3)(n+2)} \\ &= 2 + \frac{n^2-n-6}{(n+2)(n+3)}\\ &= 2 + \frac{(n-3)(n+2)}{(n+2)(n+3)} \\ &= 2 + \frac{n-3}{n+3} \\ &= \frac{2n}{n+3} \end{align*}

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1995 Paper 1 Q12
D: 1500.0 B: 1501.9
probability combinatorics arrangements permutations counting arguments gaps method

A school has \(n\) pupils, of whom \(r\) play hocket, where \(n\geqslant r\geqslant2.\) All \(n\) pupils are arranged in a row at random.

  1. What is the probability that there is a hockey player at each end of the row?
  2. What is the probability that all the hockey players are standing together?
  3. By considering the gaps between the non-hockey-players, find the probability that no two hockey players are standing together, distinguishing between cases when the probability is zero and when it is non-zero.

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