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1990 Paper 3 Q15
D: 1700.0 B: 1482.6
probability geometric distribution marginal distribution independence joint distribution expected value summation discrete random variables

An unbiased twelve-sided die has its faces marked \(A,A,A,B,B,B,B,B,B,B,B,B.\) In a series of throws of the die the first \(M\) throws show \(A,\) the next \(N\) throws show \(B\) and the \((M+N+1)\)th throw shows \(A\). Write down the probability that \(M=m\) and \(N=n\), where \(m\geqslant0\) and \(n\geqslant1.\) Find

  1. the marginal distributions of \(M\) and \(N\),
  2. the mean values of \(M\) and \(N\).
Investigate whether \(M\) and \(N\) are independent. Find the probability that \(N\) is greater than a given integer \(k\), where \(k\geqslant1,\) and find \(\mathrm{P}(N > M).\) Find also \(\mathrm{P}(N=M)\) and show that \(\mathrm{P}(N < M)=\frac{1}{52}.\)

Solution: \begin{align*} \mathbb{P}(M = m, N = n) &= \left ( \frac{3}{12} \right)^m \left ( \frac{9}{12} \right)^n \frac{3}{12} \\ &= \frac{3^n}{4^{m+n+1}} \end{align*}

  1. \begin{align*} \mathbb{P}(M = m) &= \sum_{n = 1}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{n = 1}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{1}{4^{m+1}} \sum_{n = 1}^{\infty} \left ( \frac34\right)^n \\ &= \frac{1}{4^{m+1}} \frac{3/4}{1/4} \\ &= \frac{3}{4^{m+1}} \\ \\ \mathbb{P}(N = n) &= \sum_{m = 0}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{m = 0}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{3^n}{4^{n+1}} \sum_{m = 0}^{\infty} \left ( \frac14\right)^n \\ &= \frac{3^n}{4^{n+1}} \frac{1}{3/4} \\ &= \frac{3^{n-1}}{4^{n}} \\ \end{align*}
  2. \(M+1 \sim Geo(\frac34) \Rightarrow \mathbb{E}(M) = \frac43 -1 = \frac13\) \(N \sim Geo(\frac14) \Rightarrow \mathbb{E}(N) = 4\)
\(M,N\) are independent since \(\mathbb{P}(M = m, N =n ) = \mathbb{P}(M=m)\mathbb{P}(N=n)\) \begin{align*} \mathbb{P}(N > k) &= \sum_{n=k+1}^{\infty} \mathbb{P}(N = n) \\ &= \sum_{n=k+1}^{\infty} \frac{3^{n-1}}{4^{n}} \\ &= \frac{3^k}{4^{k+1}} \sum_{n = 0}^{\infty} \left ( \frac34\right)^n \\ &= \frac{3^k}{4^{k+1}} \frac{1}{1/4} \\ &= \frac{3^k}{4^k} \end{align*} \begin{align*} \mathbb{P}(N > M) &= \sum_{m=0}^{\infty} \mathbb{P}(N > m) \mathbb{P}(M = m) \\ &= \sum_{m=0}^{\infty} \left (\frac34 \right)^m \frac{3}{4^{m+1}}\\ &=\sum_{m=0}^{\infty} \frac{3^{m+1}}{4^{2m+1}}\\ &= \frac{3}{4} \frac{1}{13/16} \\ &= \frac{12}{13} \\ \\ \mathbb{P}(N=M) &= \sum_{m=1}^{\infty} \mathbb{P}(N=m, M=m) \\ &= \sum_{m=1}^{\infty} \frac{3^m}{4^{2m+1}} \\ &= \frac{3}{64} \sum_{m=0}^{\infty} \left ( \frac{3}{16} \right)^m \\ &= \frac{3}{64} \frac{1}{13/16} \\ &= \frac{3}{52}\\ \\ \mathbb{P}(N < M) &= 1 - \frac34 - \frac3{52} \\ &= 1 - \frac{48}{52} - \frac{3}{52} \\ &= 1 - \frac{51}{52} \\ &= \frac{1}{52} \end{align*}

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1987 Paper 1 Q14
D: 1500.0 B: 1500.0
probability geometric distribution sequential games tournament conditional probability infinite series combinatorics

\(A,B\) and \(C\) play a table tennis tournament. The winner of the tournament will be the first person to win two games in a row. In any game, whoever is not playing acts as a referee, and each playerhas equal chance of winning the game. The first game of the tournament is played between \(A\) and \(B\), with \(C\) as referee. Thereafter, if the tournament is still undecided at the end of any game, the winner and referee of that game play the next game. The tournament is recorded by listing in order the winners of each game, so that, for example, \(ACC\) records a three-game tournament won by \(C\), the first game having been won by \(A\). Determine which of the following sequences of letters could be the record of a complete tournament, giving brief reasons for your answers:

  1. \(ACB\),
  2. \(ABB\),
  3. \(ACBB\).
Find the probability that the tournament is still undecided after 5 games have been played. Find also the probabilities that each of \(A,B\) and \(C\) wins in 5 or fewer games. Show that the probability that \(A\) wins eventually is \(\frac{5}{14}\), and find the corresponding probabilities for \(B\) and \(C\).

Solution:

  1. \(ACB\) is not a complete tournament since no-one has won two matches.
  2. \(ABB\) is not a possible complete tournament since it implies \(B\) won game 2, which is between \(A\) (winner of game 1) and \(C\) (referee of game 1).
  3. \(ACBB\) is a valid tournament, \(A\) beat \(B\), then \(C\) beat \(A\), then \(B\) beat \(C\) and finally \(B\) beat \(A\) to win.
After the first game there is always someone playing for the tournament, so for there to be no result after 5 games, 4 games must have gone against the leader, so the probability is \(\frac{1}{2^4} = \frac{1}{16}\). If \(A\) wins their first game, they can either win in two games (WW) or in five games (WLRWW). The probability of this is \(\frac14 + \frac1{16} = \frac{5}{16}\). Similarly \(B\) has exactly the same chance as \(A\) since everything about them is symmetric, ie a probability of \(\frac5{16}\) of winning. Since there is a \(\frac{15}{16}\) chance the tournament is decided after 5 games, the remaining \(\frac{5}{16}\) must be \(C\)'s chance of winning. After the first game is played, there's \(3\) states for each player. King (about to win if they win, becomes Ref if they lose), Challenger (needs to win to become king) and Ref (who becomes Challenger if Challenger wins). \begin{align*} \P(\text{King wins}) &= \frac{1}{2} + \frac{1}{2}\P(\text{Ref wins})\\ \P(\text{Challenger wins}) &= \frac{1}{2} \P(\text{King wins}) \\ \P(\text{Ref wins}) &= \frac{1}{2} \P(\text{Challenger wins}) \\ \end{align*} \(p_K = \frac12 + \frac12 (\frac12 \frac12 p_K) \Rightarrow \frac78 p_K = \frac12 \Rightarrow p_K = \frac47, p_C = \frac27, p_R = \frac17\). \(A\) has \(\frac12\) of being king, \(\frac12\) of being ref after the first match, so \(\frac12 \frac47 + \frac12 \frac17 = \frac{5}{14}\). Similarly \(B\) has \(\frac5{14}\) chance of winning, but unfortunately \(C\) must be the challenger after the first match and only has \(\frac27 = \frac4{14}\) chances of winning.

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