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2011 Paper 1 Q12
D: 1500.0 B: 1470.2
probability conditional probability combinatorics counting arrangements ballot problem queuing change-making case analysis

I am selling raffle tickets for \(\pounds1\) per ticket. In the queue for tickets, there are \(m\) people each with a single \(\pounds1\) coin and \(n\) people each with a single \(\pounds2\) coin. Each person in the queue wants to buy a single raffle ticket and each arrangement of people in the queue is equally likely to occur. Initially, I have no coins and a large supply of tickets. I stop selling tickets if I cannot give the required change.

  1. In the case \(n=1\) and \(m\ge1\), find the probability that I am able to sell one ticket to each person in the queue.
  2. By considering the first three people in the queue, show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=2\) and \(m\ge2\) is \(\dfrac{m-1}{m+1}\,\).
  3. Show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=3\) and \(m\ge3\) is \(\dfrac{m-2}{m+1}\,\).

Solution:

  1. The only way you wont be able to sell to them is if they are first, ie \(\frac1{m+1}\)
  2. If \(n=2\), the the only way you fail to sell to them is if one comes first or they both appear before two people with pound coins, ie \(2\) or \(122\). These have probabilities \(\frac{2}{m+2}\) and \(\frac{m}{m+2} \cdot \frac{2}{m+1} \frac{1}{m} = \frac{2}{(m+1)(m+2)}\). Therefore the total probability you don't sell all the tickets is \(\frac{2}{m+2}\left ( 1 + \frac{1}{m+1} \right) = \frac{2}{m+2} \frac{m+2}{m+1} = \frac{2}{m+1}\). Therefore the probability you do sell all the tickets is \(1 - \frac{2}{m+1} = \frac{m-1}{m+1}\)
  3. The only ways to fail when \(n=3\) are: \(2\), \(122\), or if all three \(2\)s appear before three \(1\)s. this can happen in \(11222\), \(12122\) These happen with probability: \begin{align*} 2: && \frac{3}{m+3} \\ 122: && \frac{m}{m+3} \cdot \frac{3}{m+2} \cdot \frac{2}{m+1} \\ 11222: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ 12122: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ \end{align*} Therefore the total probability is: \begin{align*} P &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+2)(m+1)+6m + 12 \right) \\ &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+1)(m+2) \right) \\ &= \frac{3}{m+1} \end{align*} and the result follows

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1997 Paper 2 Q14
D: 1600.0 B: 1469.6
Poisson distribution probability conditional probability expected value traffic model queuing relative velocity binomial distribution

Traffic enters a tunnel which is 9600 metres long, and in which overtaking is impossible. The number of vehicles which enter in any given time is governed by the Poisson distribution with mean 6 cars per minute. All vehicles travel at a constant speed until forced to slow down on catching up with a slower vehicle ahead. I enter the tunnel travelling at 30 m\(\,\)s\(^{-1}\) and all the other traffic is travelling at 32 m\(\,\)s\(^{-1}\). What is the expected number of vehicles in the queue behind me when I leave the tunnel? Assuming again that I travel at 30 m\(\,\)s\(^{-1}\), but that all the other vehicles are independently equally likely to be travelling at 30 m\(\,\)s\(^{-1}\) or 32 m\(\,\)s\(^{-1}\), find the probability that exactly two vehicles enter the tunnel within 20 seconds of my doing so and catch me up before I leave it. Find also the probability that there are exactly two vehicles queuing behind me when I leave the tunnel. \noindent [Ignore the lengths of the vehicles.]

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