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2006 Paper 3 Q12
Fifty times a year, 1024 tourists disembark from a cruise liner at a port. From there they must travel to the city centre either by bus or by taxi. Tourists are equally likely to be directed to the bus station or to the taxi rank. Each bus of the bus company holds 32 passengers, and the company currently runs 15 buses. The company makes a profit of \(\pounds\)1 for each passenger carried. It carries as many passengers as it can, with any excess being (eventually) transported by taxi. Show that the largest annual licence fee, in pounds, that the company should consider paying to be allowed to run an extra bus is approximately \[ 1600 \Phi(2) - \frac{800}{\sqrt{2\pi}}\big(1- \e^{-2}\big)\,, \] where \(\displaystyle \Phi(x) =\dfrac1{\sqrt{2\pi}} \int_{-\infty}^x \e^{-\frac12t^2}\d t\,\). You should not consider continuity corrections.
Solution: The the number of people being directed towards the buses (each cruise) is \(X \sim B(1024, \tfrac12) \approx N(512, 256) \approx 16Z + 512\). Therefore without an extra bus, the expected profit is \(\mathbb{E}[\min(X, 15 \times 32)]\). With the extra bus, the extra profit is \(\mathbb{E}[\min(X, 16 \times 32)]\), therefore the expected extra profit is: \(\mathbb{E}[\min(X, 16 \times 32)]-\mathbb{E}[\min(X, 15 \times 32)] = \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \) \begin{align*} \text{Expected extra profit} &= \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \\ &= \mathbb{E}[\min(16Z+512, 16 \times 32)-\min(16Z+512, 15 \times 32)] \\ &= 16\mathbb{E}[\min(Z+32, 32)-\min(Z+32, 30)] \\ &=16\int_{-\infty}^{\infty} \left (\min(Z+32, 32)-\min(Z+32, 30) \right)p_Z(z) \d z \\ &= 16 \left ( \int_{-2}^{0} (z+32-30) p_Z(z) \d z + \int_0^\infty (32-30)p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} (z+2) p_Z(z) \d z + \int_0^\infty 2p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} zp_Z(z) \d z + 2\int_{-2}^\infty p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} z \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} \d z + 2(1-\Phi(2)) \right) \\ &= 32(1-\Phi(2)) + \frac{16}{\sqrt{2\pi}} \left [ -e^{-\frac12z^2} \right]_{-2}^0 \\ &= 32(1-\Phi(2)) - \frac{16}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \end{align*} Across \(50\) different runs, this profit is \[ 1600(1-\Phi(2)) - \frac{800}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \]
1991 Paper 2 Q16
Each time it rains over the Cabbibo dam, a volume \(V\) of water is deposited, almost instanetaneously, in the reservoir. Each day (midnight to midnight) water flows from the reservoir at a constant rate \(u\) units of volume per day. An engineer, if present, may choose to alter the value of \(u\) at any midnight.
- Suppose that it rains at most once in any day, that there is a probability \(p\) that it will rain on any given day and that, if it does, the rain is equally likely to fall at any time in the 24 hours (i.e. the time at which the rain falls is a random variable uniform on the interval \([0,24]\)). The engineers decides to take two days' holiday starting at midnight. If at this time the volume of water in the reservoir is \(V\) below the top of the dam, find an expression for \(u\) such that the probability of overflow in the two days is \(Q\), where \(Q < p^{2}.\)
- For the engineer's summer holidays, which last 18 days, the reservoir is drained to a volume \(kV\) below the top of the dam and the rate of outflow \(u\) is set to zero. The engineer wants to drain off as little as possible, consistent with the requirement that the probability that the dam will overflow is less than \(\frac{1}{10}.\) In the case \(p=\frac{1}{3},\) find by means of a suitable approximation the required value of \(k\).
- Suppose instead that it may rain at most once before noon and at most once after noon each day, that the probability of rain in any given half-day is \(\frac{1}{6}\) and that it is equally likely to rain at any time in each half-day. Is the required value of \(k\) lower or higher?
Solution:
- It cannot overflow on the first day, since it is already \(V\) below the top. The only way it can overflow is if it rains both days. This will occur with probability \(p^2\). The probability it overflows therefore is the probability that bad timing hampers us, ie \(V - u(1+t_2) > 0\) where \(t_2\) is the timing of the rain on day 2 (as a fraction of a day). Ie \(t_2 < \frac{V}{u}-1\). Therefore \begin{align*} && Q &= p^2 \left (\frac{V}{u} - 1 \right) \\ \Rightarrow && u &= \frac{Vp^2}{p^2+Q} \end{align*}
- The probability the reservoir overflows during this \(18\) days is \(\mathbb{P}(\text{rains more than }k\text{ times})\). The number of times it rains (\(X\)) is \(B(18, \tfrac13)\), since \(18 \cdot \tfrac13 = 6 > 5\) a normal approximation is reasonable, ie \(X \approx N(6, 4)\). We wish to find \(k\) such that \(\mathbb{P}( X > k + 0.5) < \tfrac1{10}\) therefore \(k \approx 1.28 \cdot 2 + 6 - 0.5 \approx 8.1\) so they should set \(k\) to \(9\)
- In this case we have \(B(36, \tfrac16)\) approximated by \(B(6, 5)\) which has a larger standard deviation, therefore we need to choose a larger value for \(k\). [It turns out to actually be the same, but there's no reason to be able to expect students without a calculator to establish this]
1989 Paper 3 Q16
It is believed that the population of Ruritania can be described as follows:
- \(25\%\) are fair-haired and the rest are dark-haired;
- \(20\%\) are green-eyed and the rest hazel-eyed;
- the population can also be divided into narrow-headed and broad-headed;
- no narrow-headed person has green eyes and fair hair;
- those who are green-eyed are as likely to be narrow-headed as broad-headed;
- those who are green-eyed and broad-headed are as likely to be fair-headed as dark-haired;
- half of the population is broad-headed and dark-haired;
- a hazel-eyed person is as likely to be fair-haired and broad-headed as dark-haired and narrow-headed.
Solution:
