Problems

Filters
Clear Filters

5 problems found

2018 Paper 2 Q13
D: 1600.0 B: 1502.8
probability tree diagram recurrence relation Markov chain random walk circular motion symmetry pass the parcel

Four children, \(A\), \(B\), \(C\) and \(D\), are playing a version of the game `pass the parcel'. They stand in a circle, so that \(ABCDA\) is the clockwise order. Each time a whistle is blown, the child holding the parcel is supposed to pass the parcel immediately exactly one place clockwise. In fact each child, independently of any other past event, passes the parcel clockwise with probability \(\frac{1}{4}\), passes it anticlockwise with probability \(\frac{1}{4}\) and fails to pass it at all with probability \(\frac{1}{2}\). At the start of the game, child \(A\) is holding the parcel. The probability that child \(A\) is holding the parcel just after the whistle has been blown for the \(n\)th time is \(A_n\), and \(B_n\), \(C_n\) and \(D_n\) are defined similarly.

  1. Find \(A_1\), \(B_1\), \(C_1\) and \(D_1\). Find also \(A_2\), \(B_2\), \(C_2\) and \(D_2\).
  2. By first considering \(B_{n+1}+D_{n+1}\), or otherwise, find \(B_n\) and \(D_n\). Find also expressions for \(A_n\) and \(C_n\) in terms of \(n\).

Solution:

  1. \(\,\) \begin{align*} && A_1 &= \frac12 \\ && B_1 &= \frac14 \\ && C_1 &= 0 \\ && D_1 &= \frac14 \end{align*} \begin{align*} && A_2 &= \frac12 \cdot \frac12 + 2 \cdot \frac14 \cdot \frac14 = \frac38 \\ && B_2 &= \frac14 \cdot \frac12 + \frac12 \cdot \frac14 = \frac14 \\ && C_2 &=2 \cdot \frac14 \cdot \frac14 =\frac18 \\ && D_2 &= B_2 = \frac14 \end{align*}
  2. \begin{align*} && A_{n+1} &= \frac12 A_n+ \frac14(B_n + D_n) \\ && B_{n+1} &= \frac12 B_n+ \frac14(A_n + C_n) \\ && C_{n+1} &= \frac12 C_n+ \frac14(D_n +B_n) \\ && D_{n+1} &= \frac12 D_n+ \frac14(C_n +A_n) \\ \\ \Rightarrow && B_{n+1}+D_{n+1} &= \frac12 (B_n+D_n) + \frac12(A_n+C_n) \\ &&&= \frac12 \\ \Rightarrow && B_{n+1}&=D_{n+1} = \frac14 \\ \\ && C_{n+1} &= \frac12C_n + \frac14 \cdot \frac12 \\ &&&= \frac12 C_n + \frac18\\ &&&= \frac12 C_{n-1} + \frac1{8} + \frac1{16} \\ &&&= \frac1{8} + \frac{1}{16} + \cdots + \frac{1}{8 \cdot 2^{n-1}} \\ &&&= \frac18 \left (1 + \frac12 + \cdots + \frac1{2^{n-1}} \right) \\ &&&= \frac18\left ( \frac{1-\frac1{2^n}}{1-\frac12} \right) \\ &&&= \frac18 \left (2 - \frac{1}{2^{n-1}} \right) \\ &&&= \frac14 - \frac{1}{2^{n-1}} \\ \Rightarrow && A_n &= \frac14 + \frac1{2^{n-1}} \end{align*}

View
2015 Paper 2 Q12
D: 1600.0 B: 1500.0
probability conditional probability tree diagram sequences Markov chain recurrence relation coin tossing Penney's game

Four players \(A\), \(B\), \(C\) and \(D\) play a coin-tossing game with a fair coin. Each player chooses a sequence of heads and tails, as follows: Player A: HHT; Player B: THH; Player C: TTH; Player D: HTT. The coin is then tossed until one of these sequences occurs, in which case the corresponding player is the winner.

  1. Show that, if only \(A\) and \(B\) play, then \(A\) has a probability of \(\frac14\) of winning.
  2. If all four players play together, find the probabilities of each one winning.
  3. Only \(B\) and \(C\) play. What is the probability of \(C\) winning if the first two tosses are TT? Let the probabilities of \(C\) winning if the first two tosses are HT, TH and HH be \(p\), \(q\) and \(r\), respectively. Show that \(p=\frac12 +\frac12q\). Find the probability that \(C\) wins.

Solution:

  1. The only way \(A\) can win is if the sequence starts HH, if it does not start like this, then the only way HHT can appear is after a sequence of THH...H, but then THH has already appeared and \(B\) has won. Therefore the probability is \(\frac14\)
  2. If HH appears before TT then either \(A\) or \(B\) will win. If HH appears first, then \(A\) has a \(\frac14\) probability of winning. So \(A\): \(\frac18\), \(B:\), \(\frac38\), \(C:\), \(\frac18\), \(D: \frac38\)
  3. If the first two tosses are TT then \(C\) will win. If the first two tosses are HT, then either the next toss is T and \(C\) wins, or the next toss is H, and it's as if we started TH. ie \(p = \frac12 + \frac12 q\). If the first two tosses are TH, then either the next toss is H and \(C\) losses or the next toss is T and it's like starting HT. So \(q = \frac12 p\). Therefore \(p = \frac12 + \frac14p \Rightarrow p = \frac13\) If the first two tosses are HH, then eventually a T appears, and it's the same as starting HT. Therefore the probability \(C\) wins is: \(\frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16 + \frac14 \cdot \frac13 = \frac{11}{24}\)

View
2003 Paper 3 Q13
D: 1700.0 B: 1500.0
probability conditional probability recurrence relation Markov chain symmetry simultaneous equations random walk

In a rabbit warren, underground chambers \(A, B, C\) and \(D\) are at the vertices of a square, and burrows join \(A\) to \(B\), \ \(B\) to \(C\), \ \(C\) to \(D\) and \(D\) to \(A\). Each of the chambers also has a tunnel to the surface. A rabbit finding itself in any chamber runs along one of the two burrows to a neighbouring chamber, or leaves the burrow through the tunnel to the surface. Each of these three possibilities is equally likely. Let \(p_A\,\), \(p_B\,\), \(p_C\) and \(p_D\) be the probabilities of a rabbit leaving the burrow through the tunnel from chamber \(A\), given that it is currently in chamber \(A, B, C\) or \(D\), respectively.

  1. Explain why \(p_A = \frac13 + \frac13p_B + \frac13 p_D\).
  2. Determine \(p_A\,\).
  3. Find the probability that a rabbit which starts in chamber \(A\) does not visit chamber~\(C\), given that it eventually leaves the burrow through the tunnel in chamber \(A\).

View
2001 Paper 3 Q13
D: 1700.0 B: 1500.0
probability conditional probability sequential game fair coin recurrence relation Markov chain non-transitive game theory

In a game for two players, a fair coin is tossed repeatedly. Each player is assigned a sequence of heads and tails and the player whose sequence appears first wins. Four players, \(A\), \(B\), \(C\) and \(D\) take turns to play the game. Each time they play, \(A\) is assigned the sequence TTH (i.e.~Tail then Tail then Head), \(B\) is assigned THH, \(C\) is assigned HHT and \(D\) is assigned~HTT.

  1. \(A\) and \(B\) play the game. Let \(p_{\mathstrut\mbox{\tiny HH}}\), \(p_{\mathstrut\mbox{\tiny HT}}\), \(p_{\mathstrut\mbox{\tiny TH}}\) and \(p_{\mathstrut\mbox{\tiny TT}}\) be the probabilities of \(A\) winning the game given that the first two tosses of the coin show HH, HT, TH and TT, respectively. Explain why \(p_{\mathstrut\mbox{\tiny TT}} = 1\,\), and why $p_{\mathstrut\mbox{\tiny HT}} = {1 \over 2} \, p_{\mathstrut\mbox{\tiny TH}} + {1\over 2} \, p_{\mathstrut\mbox{\tiny TT}}\,$. Show that $p_{\mathstrut\mbox{\tiny HH}} = p_{\mathstrut\mbox{\tiny HT}} = {2 \over 3}$ and that \(p_{\mathstrut\mbox{\tiny TH}} = {1\over 3}\,\). Deduce that the probability that A wins the game is \({2\over 3}\,\).
  2. \(B\) and \(C\) play the game. Find the probability that \(B\) wins.
  3. Show that if \(C\) plays \(D\), then \(C\) is more likely to win than \(D\), but that if \(D\) plays \(A\), then \(D\) is more likely to win than \(A\).

View
1996 Paper 3 Q12
D: 1700.0 B: 1554.3
conditional probability Bayes' theorem Markov chain transition matrix stationary distribution simultaneous equations law of total probability

It has been observed that Professor Ecks proves three types of theorems: 1, those that are correct and new; 2, those that are correct, but already known; 3, those that are false. It has also been observed that, if a certain of her theorems is of type \(i\), then her next theorem is of type \(j\) with probability \(p\low_{ij},\) where \(p\low_{ij}\) is the entry in the \(i\)th row and \(j\)th column of the following array: \[ \begin{pmatrix}0.3 & 0.3 & 0.4\\ 0.2 & 0.4 & 0.4\\ 0.1 & 0.3 & 0.6 \end{pmatrix}\,. \] Let \(a_{i},\) \(i=1,2,3\), be the probability that a given theorem is of type \(i\), and let \(b_{j}\) be the consequent probability that the next theorem is of type \(j\).

  1. Explain why \(b_{j}=a\low_{1}p\low_{1j}+a\low_{2}p\low_{2j}+a\low_{3}p\low_{3j}\,.\)
  2. Find values of \(a\low_{1},a\low_{2}\) and \(a\low_{3}\) such that \(b_{i}=a_{i}\) for \(i=1,2,3.\)
  3. For these values of the \(a_{i}\) find the probabilities \(q\low_{ij}\) that, if a particular theorem is of type \(j\), then the \textit{preceding }theorem was of type \(i\).

View