Problems

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Solution: \(AG = r\), therefore the area is: \begin{align*} A &= [AHG] - 2*[ABC] + [CDE] \\ &= \frac12 \pi r^2 - \pi h^2 + \frac12 \pi (r-2h)^2 \\ &= \frac12 \pi \l r^2 - 2h^2 + r^2 -4rh+4h^2 \r \\ &= \frac12 \pi \l 2r^2 -4rh + 2h^2\r \\ &= \pi (r-h)^2 \end{align*} This is the same area as a circle radius \(r-h\) But \(HD = r + (r-2d) = 2(r-d)\), ie the circle with diameter \(HD\) has radius \(r-h\) as required. Suppose \(A = (-h, 0), C = (h, 0), B = (0, h)\) then our parabola is \(y = \frac1{h}(h^2-x^2)\)

The area of \(ABC\) would then be \(\int_{-h}^h \frac{1}{h}(h^2-x^2) \d x = \frac1{h} \left [ h^2x - \frac{x^3}{3} \right] = \frac1{h} \l 2h^3-2\frac{h^3}{3} \r = \frac{4}{3}h^2\) so we have: \begin{align*} A &= [AHG] - 2*[ABC] +[CDE] \\ &= \frac43 r^2-\frac83h^2+\frac43(r-2h)^2 \\ &= \frac43 \l r^2 -2h^2+r^2-4rh+4h^2) \\ &= \frac43 (r-h)^2 \end{align*}

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Solution: Integrating by parts we obtain: \begin{align*} \int x \ln x \, \d x &= [\frac12 x^2 \ln x] - \int \frac12x^2 \cdot \frac1x \d x \\ &= \frac12 x^2 \ln x - \frac14 x^2 + C \end{align*}

We should have: \begin{align*} \int_0^1 x \ln \frac{1}{x} \d x &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \left [ -\frac12 x^2 \ln x + \frac14 x^2 \right]_0^1 &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \frac{1}{4} &=\lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^n \l i \ln n - i \ln i \r \\ &= \lim_{n \to \infty} \frac{1}{n^2}\l \frac{n(n+1)}{2} \ln n - \sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n \sum_{k=1}^i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \sum_{i=0}^k \ln (n-i) \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \ln \frac{n!}{(n-k)!}\r \\ \end{align*}

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Solution:

Denoting \(\overrightarrow{OY}\) by \(\mathbf{y}\) and \(\overrightarrow{OC}\) by \(\mathbf{c}\) etc, we have: \begin{align*} \mathbf{c} &= \frac23 \mathbf{a} + \frac13 \mathbf{b} \\ \mathbf{d} &= \frac13 \mathbf{a} + \frac23 \mathbf{b} \\ \mathbf{x} &= \lambda \mathbf{a} \\ \mathbf{y} &= (1-\lambda) \mathbf{b} \\ 0 &= (\mathbf{d}-\mathbf{x}) \cdot (\mathbf{c} - \mathbf{y}) \\ &=((\frac13 -\lambda)\mathbf{a} + \frac23 \mathbf{b})\cdot(\frac23 \mathbf{a} + (\frac13-1+\lambda) \mathbf{b} ) \\ &= \frac{2}{3} \cdot (\frac13-\lambda) +\frac23 \cdot(\lambda - \frac23)+(\frac{4}{9}+(\frac13-\lambda)(-\frac23+\lambda))\mathbf{a}\cdot\mathbf{b} \\ &= -\frac{2}{9} + (\frac{4}{9} - \frac{2}{9}+\lambda-\lambda^2)\mathbf{a}\cdot \mathbf{b} \\ &= - \frac{2}{9} + (\frac{2}{9} + \lambda - \lambda^2)\mathbf{a}\cdot \mathbf{b} \\ \mathbf{a}\cdot \mathbf{b} &= \frac{2/9}{2/9+\lambda - \lambda^2} \end{align*} Since \(0 \leq \lambda - \lambda^2 \leq \frac14\), \(\frac{\frac29}{\frac29+\frac14} = \frac8{17} \leq \mathbf{a}\cdot\mathbf{b} \leq 1\) If \(\angle AOB\) is as large as possible, \(\mathbf{a}\cdot\mathbf{b}\) is as small as possible, ie \(\lambda = \frac12\) and \(\mathbf{a}\cdot \mathbf{b} = \frac{8}{17}\) First notice that the length \(OM\) to the midpoint of \(AB\) is \(\sqrt{\frac14 (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})} = \sqrt{\frac14 (2 + 2\mathbf{a}\cdot \mathbf{b})} = \sqrt{\frac12 + \frac4{17}} = \sqrt{\frac{25}{34}} = \frac{5}{\sqrt{34}}\) Notice that \(XYE\) and \(DCE\) are similar triangles, and so the heights satisfy \(\frac{h_1}{h_2} = \frac{\frac12}{\frac13} = \frac32\). Therefore the length \(OE\) is \(\frac12 \frac{5}{\sqrt{34}} + \frac{3}{5} \frac12 \frac{5}{\sqrt{34}} = \frac{8}{10} \frac{5}{\sqrt{34}} = \frac{4}{\sqrt{34}}\)

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Solution: Consider the \(5\) segements radiating from \(A\). By the pigeonhole principle, at least \(3\) of them must be the same colour (say gold and say reaching \(B,C,D\)). If any of the segments joining any of \(B,C,D\) are gold then we have found a monochromatic gold triangle. But if none of them are gold, they are all silver, therefore \(BCD\) is a monochromatic silver triangle.

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Solution: \[ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \]

1. \begin{align*} (1+1)^n &= \sum_{k=0}^n \binom{n}{k} \\ (1-1)^n &= \sum_{k=0}^n (-1)^n\binom{n}{k} \\ &= \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} -\sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} \end{align*} Therefore \(\displaystyle \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} = \sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} = \frac{2^n}{2} = 2^{n-1}\)
2. \begin{align*} && (1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{k=0}^n k\binom{n}{k} x^{k-1} \\ x = 1: && n2^{n-1} &= \sum_{k=1}^n k\binom{n}{k} \end{align*} as required

\begin{align*} \sum_{r=0}^n (r + (-1)^r) \binom{n}{r} &= n2^{n-1}+0 = n2^{n-1} \end{align*}

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Solution: The normal to the curve \(y = f(x)\) has gradient \(-\frac{1}{f'(x)}\) and so has equation: \begin{align*} && \frac{Y - f(x)}{X - x} &= -\frac{1}{f'(x)} \\ \Rightarrow && Y &= -\frac{1}{f'(x)}X + \frac{x}{f'(x)}+f(x) \end{align*} Hence the \(Q\) is \(\displaystyle \left (0, f(x) + \frac{x}{f'(x)} \right)\). \begin{align*} && |PQ|^2 &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && x^2 + e^{x^2} &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && (f'(x))^2 &=x^2 e^{-x^2} \end{align*} Therefore \(f'(x) = \pm x e^{-x^2/2}\). If \(f(x)\) has a minimum at \((0,-2)\) then \(f''(0) > 0\), and \(f''(x) = \pm (e^{-x^2/2} - x^2e^{-x^2/2}) = \pm e^{-x^2/2}(1-x^2)\) so we should take the positive branch of the solution, ie \(f'(x) = xe^{-x^2/2}\). Therefore \(f(x) = - e^{-x^2/2}+C\). Since \(f(0) = -2\) we must have \(-2 = -1 + C\), ie \(C = -1\). Therefore \(f(x) = -1 - e^{-x^2/2}\)

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Solution:

Suppose one of the \(x\) coordinates is \(t > 0\), then the coordinates are \(y = \pm \sqrt{1-t}, x = \pm t\). The area will be \(A = 2t \cdot 2 \sqrt{1-t}\). To maximise this, \begin{align*} && \frac{\d A}{\d t} &= 4 \sqrt{1-t} - 2t(1-t)^{-\frac12} \\ &&&= \frac{4(1-t) - 2t}{\sqrt{1-t}} \\ &&&= \frac{4-6t}{\sqrt{1-t}} \end{align*} Therefore there is a stationary point at \(t = \frac23\). Since we know the area is \(0\) when \(t = 0, 1\) we can see this must be a maximum for the area. Therefore the area is \(\displaystyle 4 \frac23 \sqrt{1-\frac23} = \frac{8}{3\sqrt{3}} = \frac{8}{\sqrt{27}}\). For this similar problem, using a similar approach we find \(y = \pm (1- t)^{n/2m}, x = \pm t\) and so the area is \(A = 4 t \cdot (1-t)^{n/2m}\). \begin{align*} && \frac{\d A}{\d t} &= 4(1-t)^{n/2m} - 4t \frac{n}{2m} (1-t)^{\frac{n}{2m} - 1} \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4(1-t) - \frac{2n}{m} t\right) \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4 - (4 + \frac{2n}{m})t \right) \\ \end{align*} Therefore \(\displaystyle t = \frac{2m}{2m+n}\) and \(\displaystyle A = 4\cdot \frac{2m}{2m+n} \cdot (1 - \frac{2m}{2m+n})^{n/2m} = \frac{8m}{2m+n} \cdot \left ( \frac{n}{2m+n}\right)^{n/2m}\)

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Solution: Given that \(e^{i \theta} = \cos \theta + i \sin \theta\) we must have that

1. \begin{align*} \cos 4 \theta &= \textrm{Re} \l e^{i 4 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^4 \r \\ &= \cos^4 \theta - \binom{4}{2}\cos^2 \theta \sin^2 \theta +\sin^4 \theta \\ &= \cos^4 \theta - 6\cos^2 \theta (1-\cos^2 \theta) +(1-\cos^2 \theta)^2 \\ &= 8\cos^4 \theta - 8\cos^2 \theta + 1 \end{align*}
2. Similarly, \begin{align*} \cos 6 \theta &= \textrm{Re} \l e^{i 6 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^6 \r \\ &= \cos^6 \theta -\binom{6}{2}\cos^4 \theta \sin^2 \theta +\binom{6}{4} \cos^2\theta \sin^4 \theta - \sin^6 \theta \\ &= \cos^6 \theta - 15 \cos^4 \theta (1-\cos^2 \theta) + 15\cos^2 \theta (1-\cos^2\theta)^2 - (1-\cos^2 \theta)^3\\ &= 31\cos^6 \theta-45\cos^4\theta+15\cos^2\theta-1+3\cos^2 \theta-3\cos^4 \theta+\cos^6 \theta \\ &= 32 \cos^6 \theta-48\cos^4 \theta+18\cos^2 \theta-1 \end{align*}

\begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1\\ &= \frac12 (32x^6-56x^4+26x^2-1) \\ &= \frac12(32x^6-48x^4+18x^2-1-(8x^4-8x^2+1)) \end{align*} Therefore if \(x = \cos \theta\) then we are looking at solving \(\cos 6 \theta = \cos 4 \theta\). \(\cos 6 \theta - \cos 4 \theta = -2 \sin 5\theta \sin \theta = 0\). So we should be looking at \(\sin 5 \theta = 0\) and \(\sin \theta = 0\). \(\sin \theta = 0 \Rightarrow x = \cos \theta = \pm 1\) both of which are roots. The other roots will be \(\cos \frac{\pi}{5}, \cos \frac{2\pi}{5}\) etc but it's unclear this is an acceptable form. Alternatively, given our two roots, we can factorize \begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1 \\ &= (x^2-1)(16x^4-12x^2+1) \end{align*} We can solve \(16y^2-12y+1=0\) to see that \(x^2 = \frac{3 \pm \sqrt{5}}{8}\) so our roots are: \(x = -1, 1, \pm \sqrt{\frac{3 + \sqrt{5}}{8}}, \pm \sqrt{\frac{3 -\sqrt{5}}{8}}\) (We might notice that \(3+\sqrt{5} =\l \frac{1+\sqrt{5}}{\sqrt{2}} \r^2\) so our final answer could be: \(x = -1, 1, \pm \frac{1+\sqrt{5}}{4}, \pm \frac{\sqrt{5}-1}{4}\))

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Solution: \(8\frac{\d y}{\d x} = 3(x^2-4)\) so the turning points are at \((\pm 2, \mp 2)\)

We need either \begin{align*} && \begin{cases} -2a+b &= 0 \\ 2c+d &= 0 \\ 2a+b &= 1 \\ -2c+d &= 1 \end{cases} && \text{ or } && \begin{cases} -2a+b &= 1 \\ 2c+d &= 1 \\ 2a+b &= 0 \\ -2c+d &= 0 \end{cases} \\ \Rightarrow && \begin{cases} -2a+b &= 0 \\ 2a+b &= 1 \\ 2c+d &= 0 \\ -2c+d &= 1 \end{cases} && \text{ or } && \begin{cases} -2a+b &= 1 \\ 2a+b &= 0 \\ 2c+d &= 1 \\ -2c+d &= 0 \end{cases}\\ \Rightarrow && \begin{cases} (a,b) = (\frac14,\frac12) \\ (c,d) = (-\frac14, \frac12)\end{cases} && \text{ or } && \begin{cases} (a,b) = (-\frac14,\frac12) \\ (c,d) = (\frac14, \frac12)\end{cases} \end{align*} So either \(X = \frac14 x + \frac12, Y = -\frac14 y + \frac12\) or \(X = -\frac14x + \frac12, Y = \frac14y + \frac12\)

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Solution: Using Newton's second law, we have: \begin{align*} && -M\omega(v^2+V^2)/v &= M v \frac{\d v}{\d x} \\ \Rightarrow && \frac{v^2}{v^2+V^2} \frac{\d v}{\d x} &= -\omega \\ \Rightarrow && \omega X &= \int_{V/2}^V \frac{v^2}{v^2+V^2} \d v \\ &&&= \int_{V/2}^V \l 1 - \frac{V^2}{v^2+V^2} \r \d v \\ &&&= \left [v - V\tan^{-1} \frac{v}{V} \right]_{V/2}^V \\ &&&= V \l \frac12 - \tan^{-1} 1 + \tan^{-1} \frac12 \r \\ \Rightarrow X &= \frac{V}{\omega} \l \tan^{-1} \frac12 + \frac12 - \frac{\pi}{4} \r \end{align*}. The resistive force just before the field changes is \(M \omega (\frac{V^2}{4} + V^2)/\frac{V}{2} = \frac52MV\omega\). Therefor the constant resistive force is between \(\frac{11}4MV\omega\) and \(\frac{9}{4}MV \omega\) and acceleration is \(\frac{11}{4}V\omega, \frac{9}{4}V\omega\). Since \(v^2 = u^2 + 2as \Rightarrow s = \frac{v^2-u^2}{2a} = \frac{\frac{V^2}{4}}{2kV\omega} = \frac{V}{8k\omega}\) therefore \(z \in \left [ \frac{V}{22\omega},\frac{V}{18 \omega} \right]\)

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Solution: Notice that \(u_x = v \cos \theta, u_y = v \sin \theta\). We are interested in the time taken for the shot to hit the ground. \(-h = u_y t -\frac12 g t^2\) since our distance will be \(v \cos \theta \cdot t\). Solving this quadratic for \(t\) we obtain: \begin{align*} && 0 &= h + v \sin \theta \cdot t - \frac12 g \cdot t^2 \\ \Rightarrow && t_\pm &= \frac{-v \sin \theta \pm \sqrt{v^2 \sin^2 \theta+2hg}}{-g} \\ \Rightarrow && t_- &= \frac{v \sin \theta+v\sin \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}}}{g} \\ \Rightarrow && v \cos \theta t_{-} &= \frac{v^2}{g} \cos \theta \sin \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right) \\ &&&= \frac{v^2}{2g} \sin 2 \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right) \end{align*} Differentiating \(R\) wrt to \(\theta\) at \(\frac{\pi}{4}\) we obtain: \begin{align*} \frac{\d R}{\d \theta} &= \frac{v^2}{2g} \left (2 \cos 2 \theta + 2 \cos 2 \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} + \sin 2\theta \left ( 1 + \frac{2hg}{v^2 \sin^2 \theta}\right)^{-\frac12} \frac12 \frac{2hg} {v^2}(-2) \frac{\cos \theta}{\sin^3 \theta}\right) \\ \frac{\d R}{\d \theta} \biggr \rvert_{\theta = \frac{\pi}{4}} &=\frac{v^2}{2g}\left(0+0- 4\left ( 1 + \frac{4hg}{v^2 }\right)^{-\frac12} \frac{hg} {v^2} \right) \\ &< 0 \end{align*} Therefore, since \(R\) is locally decreasing in \(\theta\) he should reduce his angle of projection slightly.

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Solution:

The distance of \(A\) to \(X\) is \(\frac34\) the distance from \(A\) to the centre base (\(d\)) The distance of \(C\) to the centre of the base (\(G\)) is \(\frac{2}{3}\) the height of \(BCD\) which is \(\frac{\sqrt{3}}{2} \cdot 2a = \sqrt{3} a\). Therefore we must have \((2a)^2 = d^2 + \frac43a^2 \Rightarrow d = \frac{2\sqrt{2}}{\sqrt{3}}a\) and so \(AX = \frac34 \frac{2\sqrt{2}}{\sqrt{3}}a = \sqrt{\frac32}a\) The tension in each string will be \(\lambda \left (\sqrt{\frac32}-1 \right)\). Considering the energy of the system, when the ABCD reaches it's maximum height, it's velocity will be \(0\). Therefore the only energies to consider are GPE and EPE. Assuming the table is \(0\), we initially have \(EPE\) of \begin{align*} 3 \cdot \frac12 \lambda \frac{(a(\sqrt{\frac32}-1))^2}{a} = \frac32 \lambda a\left (\frac52-2\sqrt{\frac32} \right ) \end{align*} When \(BCD\) is level with \(O\), the height is \(\frac{1}{\sqrt{6}}a\) and GPE of \(\frac{Mga}{\sqrt{6}}\) The \(EPE\) will be: \begin{align*} 3 \cdot \frac12 \lambda \frac{(a(\frac{2}{\sqrt{3}}-1))^2}{a} &= \frac32 \lambda a \left (\frac73 - \frac{4}{\sqrt{3}}\right ) \end{align*} So by conservation of energy: \begin{align*} && \frac32 \lambda a\left (\frac52-2\sqrt{\frac32} \right ) &= \frac{Mga}{\sqrt{6}} + \frac32 \lambda a \left (\frac73 - \frac{4}{\sqrt{3}}\right ) \\ \Rightarrow && \frac{Mg}{\lambda} &= \sqrt{6} \left (\frac32 \left (\frac52-2\sqrt{\frac32} \right ) - \frac32 \left (\frac73 - \frac{4}{\sqrt{3}}\right ) \right) \\ &&&= -9 + 6\sqrt{2}+\sqrt{\frac38} \\ &&&= 0.09765380\ldots \\ &&&= 0.098\, (2\text{ s.f}) \end{align*}

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Solution: \begin{align*} \overset{\curvearrowleft}{X}: && 0&= \frac{l}{2} Mg\sin \theta - l R_1 \cos \theta \\ \Rightarrow && R_1 &= \frac12 \tan \theta Mg \\ \text{N2}(\uparrow): && 0 &= R\cos \phi +F \sin \phi - Mg \\ \text{N2}(\rightarrow):&& 0&=R_1-F \cos \phi + R \sin \phi \\ \Rightarrow && \frac12 \tan \theta Mg &= F \cos \phi- R \sin \phi \\ && Mg &= F \sin \phi +R \cos \phi \\ \Rightarrow && F &= Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) \\ && N &= Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \end{align*} If \(\mu = 2\) and \(\phi = 45^{\circ}\), we must have \(F \leq 2 N\), so: \begin{align*} && Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) &\leq 2 Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \\ \Rightarrow && 1 + \frac12 \tan \theta \leq 2-\tan \theta \\ \Rightarrow && \frac 32 \tan \theta \leq 1 \\ \Rightarrow && \tan \theta \leq \frac23 \\ \Rightarrow && \theta \leq \tan^{-1} \frac23 \end{align*}

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Solution: \begin{align*} \mathbb{P}(\text{fever tree grows}) &= \mathbb{P}(0 \leq L \leq 1) + \mathbb{P}(2 \leq L \leq 3) \\ &= \int_0^1 \frac12 -\frac18 x \d x + \int_2^3 \frac12 - \frac18 x \d x \\ &= \left [\frac12 x - \frac1{16}x^2 \right]_0^1+ \left [\frac12 x - \frac1{16}x^2 \right]_2^3 \\ &= \frac12 - \frac1{16}+\frac32-\frac9{16} - 1 + \frac{4}{16} \\ &= \frac58 \end{align*} The expected number of fever trees is just \(96 \cdot \frac58 = 60\).

1. \(f_Y(t)\) must match the distribution for \(L\), but limited to the points we care about, therefore it should be: $f_Y(t) = \begin{cases} ( \frac45 - \frac15t ) & \text{if } t \in [0,1]\cup[2,3] \\ 0 & \text{otherwise} \end{cases}$
   

   + - ↺
   \begin{align*} \mathbb{E}(Y) &= \frac12 \cdot \frac15 (4 - \frac12)+\frac52 \cdot (1 - \frac15 (4 - \frac12)) \\ &= \frac12 \cdot \frac7{10} + \frac52 \cdot \frac3{10} \\ &= \frac{22}{20} \\ &= \frac{11}{10} \end{align*}
2. Given the seeds are blown independently and the wind hasn't changed, it is reasonable to model the number of fever trees as \(B(96, \frac{5}{8})\), it is also acceptable to approximate this using a Normal distribution, ie \(N(60, 22.5)\), \(23\) is \(\frac{23-60}{\sqrt{22.5}}\) is a very negative number, so he should be extremely suspicious.

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Solution: \(A \sim 5\cdot 5 + U[0,10]\) \(B \sim 4 \cdot 5 + 3 \textrm{Po}(4)\) \(C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5\) \begin{align*} && \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\ && \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\ &&&= \mathbb{P}(Po(4) \leq 2) \\ &&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\ &&&= 0.23810\ldots \\ && \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\ &&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\ &&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\ &&&= 0.26272 \end{align*} \begin{align*} \mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\ &= \frac{\frac13 \cdot 0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot 0.23810\ldots + \frac13 \cdot 0.26272} \\ &= 0.3397\ldots \\ &= 0.340 \, \, (3\text{ s.f.}) \end{align*}

1. \begin{align*} \mathbb{E}(A) &= 25 + 5 &= 30 \\ \mathbb{E}(B) &= 20 + 3\cdot 4 &= 32 \\ \mathbb{E}(C) &= 10 + 5 \cdot 4 &= 30 \end{align*} \(A\) and \(C\) are equally good.
2. \begin{align*} \mathbb{P}(A \leq 32) &= \mathbb{P}(U \leq 7) &= 0.7 \\ \mathbb{P}(B \leq 32) &= \mathbb{P}(Po(4) \leq 4) \\ &= e^{-4}(1 + 4 + 8 + \frac{4^3}{6}) &= 0.4334\ldots \\ \mathbb{P}(C \leq 32) &= \mathbb{P}(B(5,\tfrac45) \leq 4) \\ &= 1 - \mathbb{P}(B(5,\tfrac45) = 5) \\ &= 1 - \frac{4^5}{5^5} &=0.67232 \end{align*} So you should choose route \(A\).