Problems

Filters
Clear Filters

2 problems found

2013 Paper 2 Q13
D: 1600.0 B: 1516.0
probability geometric distribution biased coin runs inequality sequences combinatorial probability

A biased coin has probability \(p\) of showing a head and probability \(q\) of showing a tail, where \(p\ne0\), \(q\ne0\) and \(p\ne q\). When the coin is tossed repeatedly, runs occur. A straight run of length \(n\) is a sequence of \(n\) consecutive heads or \(n\) consecutive tails. An alternating run of length \(n\) is a sequence of length \(n\) alternating between heads and tails. An alternating run can start with either a head or a tail. Let \(S\) be the length of the longest straight run beginning with the first toss and let \(A\) be the length of the longest alternating run beginning with the first toss.

  1. Explain why \(\P(A=1)=p^2+q^2\) and find \(\P(S=1)\). Show that \(\P(S=1)<\P(A=1)\).
  2. Show that \(\P(S=2)= \P(A=2)\) and determine the relationship between \(\P(S=3)\) and \( \P(A=3)\).
  3. Show that, for \(n>1\), \(\P(S=2n)>\P(A=2n)\) and determine the corresponding relationship between \(\P(S=2n+1)\) and \(\P(A=2n+1)\). [You are advised not to use \(p+q=1\) in this part.]

Solution:

  1. The only way \(A = 1\) is if we get \(HH\) or \(TT\) which has probability \(p^2+q^2\). The only way we get \(S=1\) is if we have \(HT\) to \(TH\), ie \(2pq\). Since \((p-q)^2 = p^2 + q^2 - 2pq >0\) we must have \(\mathbb{P}(A=1) > \mathbb{P}(S=1)\).
  2. \(\,\) \begin{align*} \mathbb{P}(S=2) &= p^2q + q^2p \\ \mathbb{P}(A=2) &= pq^2 + qp^2 = \mathbb{P}(S=2) \\ \\ \mathbb{P}(S=3) &= p^3q + q^3p = pq(p^2+q^2) \\ \mathbb{P}(A=3) &= pqp^2 + qpq^2 = pq(p^2+q^2) = \mathbb{P}(S=3) \end{align*}
  3. For \(n > 1\) we must have \begin{align*} && \mathbb{P}(S = 2n) &= p^{2n}q + q^{2n}p \\ && \mathbb{P}(A=2n) &= (pq)^{n}q + (qp)^{n}p \\ &&&= p^nq^{n+1} + q^np^{n+1} \\ && \mathbb{P}(S = 2n) &> \mathbb{P}(A = 2n) \\ \Leftrightarrow && p^{2n}q + q^{2n}p & > p^nq^{n+1} + q^np^{n+1}\\ \Leftrightarrow && 0 & < p^{2n}q+q^{2n}p - p^nq^{n+1} -q^np^{n+1}\\ &&&= (p^n-q^n)(qp^n - pq^n) \end{align*} which is clearly true. \begin{align*} && \mathbb{P}(S=2n+1) &= p^{2n+1}q + q^{2n+1}p \\ && \mathbb{P}(A=2n+1) &= (pq)^{n}p^2 + (qp)^{n}q^2 \\ &&&= p^{n+2}q^n + q^{n+2}p^n \end{align*} The same factoring logic shows that \(\mathbb{P}(S = 2n+1) > \mathbb{P}(A=2n+1)\)

View
1998 Paper 1 Q13
D: 1484.0 B: 1532.0
conditional probability Bayes' theorem sampling with replacement combinatorial probability geometric series

I have a bag initially containing \(r\) red fruit pastilles (my favourites) and \(b\) fruit pastilles of other colours. From time to time I shake the bag thoroughly and remove a pastille at random. (It may be assumed that all pastilles have an equal chance of being selected.) If the pastille is red I eat it but otherwise I replace it in the bag. After \(n\) such drawings, I find that I have only eaten one pastille. Show that the probability that I ate it on my last drawing is \[\frac{(r+b-1)^{n-1}}{(r+b)^{n}-(r+b-1)^{n}}.\]

View