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2006 Paper 1 Q12
D: 1500.0 B: 1499.3
probability conditional probability independence factorisation limiting cases network reliability

Oxtown and Camville are connected by three roads, which are at risk of being blocked by flooding. On two of the three roads there are two sections which may be blocked. On the third road there is only one section which may be blocked. The probability that each section is blocked is \(p\). Each section is blocked independently of the other four sections. Show that the probability that Oxtown is cut off from Camville is \(p^3 \l 2-p \r^2\). I want to travel from Oxtown to Camville. I choose one of the three roads at random and find that my road is not blocked. Find the probability that I would not have reached Camville if I had chosen either of the other two roads. You should factorise your answer as fully as possible. Comment briefly on the value of this probability in the limit \(p\to1\).

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2005 Paper 1 Q5
D: 1484.0 B: 1528.7
integration power rule logarithm limiting cases integration by parts approximation definite integral

  1. Evaluate the integral \[ \int_0^1 \l x + 1 \r ^{k-1} \; \mathrm{d}x \] in the cases \(k\ne0\) and \(k = 0\,\). Deduce that \(\displaystyle \frac{2^k - 1}{k} \approx \ln 2\) when \(k \approx 0\,\).
  2. Evaluate the integral \[ \int_0^1 x \l x + 1 \r ^m \; \mathrm{d}x \; \] in the different cases that arise according to the value of \(m\).

Solution:

  1. Case \(k \neq 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \left [\frac{1}{k}(x+1)^k \right]_0^1 \\ &&&= \frac{2^k-1}{k} \\ \end{align*} Case \(k = 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \int_0^1 (x+1)^{-1} \d x \\ &&&= \left [\ln(x+1) \right]_0^1 \\ &&&= \ln 2 \end{align*} Therefore for \(k \approx 0\), we must have both integrals being close to each other, since the function is nice on this interval, ie \(\frac{2^k-1}{k} \approx \ln 2\)
  2. Case \(m = 0\). \(I = \frac12\) Case \(m \neq 0, -1, -2\) \begin{align*} u = x+1, \d u = \d x && \int_0^1 x(x+1)^m \d x &= \int_{u=1}^{u=2} (u-1)u^m \d u \\ &&&=\left[ \frac{u^{m+2}}{m+2} - \frac{u^{m+1}}{m+1} \right]_1^2 \\ &&&= 2^{m+1}\left ( \frac{2}{m+2} - \frac1{m+1} \right) - \frac{1}{m+2} + \frac{1}{m+1} \\ &&&= 2^{m+1} \frac{m}{(m+1)(m+2)} + \frac{1}{(m+1)(m+2)} \\ &&&= \frac{m2^{m+1}+1}{(m+1)(m+2)} \\ \end{align*} Case \(m = -1\). \begin{align*} && \int_0^1 \frac{x}{x+1} \d x &= \int_0^1 1 - \frac{1}{x+1} \d x \\ &&&= 1 - \ln2 \\ \end{align*} Case \(m = -2\): \begin{align*} && \int_0^1 \frac{x}{(x+1)^2} \d x &= \int_0^1\frac{x+1-1}{(x+1)^2} \d x \\ &&&= \left [ \ln (x+1) +(1+x)^{-1} \right]_0^1 \\ &&&= \ln 2 + \frac12 - 1 \\ &&&= \ln 2 - \frac12 \end{align*}

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