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2016 Paper 1 Q11
D: 1516.0 B: 1484.7

The point \(O\) is at the top of a vertical tower of height \(h\) which stands in the middle of a large horizontal plain. A projectile \(P\) is fired from \(O\) at a fixed speed \(u\) and at an angle \(\alpha\) above the horizontal. Show that the distance \(x\) from the base of the tower when \(P\) hits the plain satisfies \[ \frac{gx^2}{u^2} = h(1+\cos 2\alpha) + x \sin 2\alpha \,. \] Show that the greatest value of \(x\) as \(\alpha\) varies occurs when \(x=h\tan2\alpha\) and find the corresponding value of \(\cos 2\alpha\) in terms of \(g\), \(h\) and \(u\). Show further that the greatest achievable distance between \(O\) and the landing point is \(\dfrac {u^2}g +h\,\).

Solution: \begin{align*} \rightarrow: && x &= u \cos \alpha t\\ \Rightarrow && t &= \frac{x}{u \cos \alpha}\\ \uparrow: && -h &= u\sin \alpha t- \frac12gt^2 \\ && - h &= x\tan \alpha - \frac12 g \frac{x^2}{u^2}\sec^2 \alpha \\ \Rightarrow && \frac{gx^2}{u^2} &= h(2\cos^2 \alpha) + x2 \tan \alpha \cos^2 \alpha \\ &&&= h(1 + \cos 2 \alpha) + x \sin 2\alpha \\ \frac{\d}{\d \alpha}: && \frac{g}{u^2} 2 x \frac{\d x}{\d \alpha} &= -2h \sin 2 \alpha + 2x \cos 2 \alpha +\frac{\d x}{\d \alpha} \sin 2 \alpha \\ \Rightarrow && \frac{\d x}{\d \alpha} \left ( \frac{2xg}{u^2} - \sin 2 \alpha \right) &= 2\cos 2 \alpha (x -h \tan 2 \alpha) \end{align*} Since the turning point will be a maximum must be \(x = h \tan 2 \alpha\). Therefore, let \(c = \cos 2 \alpha\) \begin{align*} && \frac{gh^2}{u^2} \tan^2 2 \alpha &= h(1 + \cos 2 \alpha) + h \tan 2 \alpha \sin 2 \alpha \\ \Rightarrow && \frac{gh}{u^2}(c^{-2}-1) &= 1+c+\frac{1-c^2}{c} \\ \Rightarrow && \frac{gh(1-c^2)}{u^2c^2} &= \frac{c+c^2+1-c^2}{c}\\ &&&= \frac{1+c}{c} \\ \Rightarrow && \frac{gh(1-c)}{u^2c} &= 1 \\ \Rightarrow && u^2c &= gh(1-c) \\ \Rightarrow && c(u^2+gh) &= gh \\ \Rightarrow && \cos 2 \alpha &= \frac{gh}{u^2+gh} \\ \\ \Rightarrow && d_{max}^2 &= h^2 + h^2 \tan^2 2 \alpha \\ &&&= h^2\sec^2 2 \alpha \\ &&&= h^2 \frac{(u^2+gh)^2}{g^2h^2} \\ &&&= \frac{(u^2+gh)^2}{g^2} \\ &&&= \left (\frac{u^2}{g}+h \right)^2 \\ \Rightarrow && d_{max} &= \frac{u^2}{g}+h \end{align*}

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2016 Paper 1 Q12
D: 1516.0 B: 1484.7

  1. Alice tosses a fair coin twice and Bob tosses a fair coin three times. Calculate the probability that Bob gets more heads than Alice.
  2. Alice tosses a fair coin three times and Bob tosses a fair coin four times. Calculate the probability that Bob gets more heads than Alice.
  3. Let \(p_1\) be the probability that Bob gets the same number of heads as Alice, and let~\(p_2\) be the probability that Bob gets more heads than Alice, when Alice and Bob each toss a fair coin \(n\) times. Alice tosses a fair coin \(n\) times and Bob tosses a fair coin \(n+1\) times. Express the probability that Bob gets more heads than Alice in terms of \(p_1\) and \(p_2\), and hence obtain a generalisation of the results of parts (i) and (ii).

Solution:

  1. There are several possibilities \begin{array}{c|c|c} \text{Alice} & \text{Bob} & P \\ \hline 0 & 1 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 2 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ 1 & 2 & 2 \cdot \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{6}{2^5} \\ 1 & 3 & 2\cdot \frac1{2^2} \cdot \frac{1}{2^3} = \frac{2}{2^5} \\ 2 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ \hline && \frac{1}{2^5}(3+3+1+6+2+1) = \frac{16}{2^5} = \frac12 \end{array}
  2. There are several possibilities \begin{array}{c|c|c} A & B & \text{count} \\ \hline 0 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 3 & 4 \\ 0 & 4 & 1 \\ 1 & 2 & 3\cdot6 \\ 1 & 3 & 3\cdot4 \\ 1 & 4 & 3 \\ 2 & 3 & 3\cdot4 \\ 2 & 4 & 3 \\ 3 & 4 & 1 \\ \hline && 64 \end{array} Therefore the total probability is \(\frac12\)
  3. \(\mathbb{P}(\text{Bob more than Alice}) = p_1 \cdot \underbrace{\frac12}_{\text{he wins by breaking the tie on his last flip}} + p_2\) If \(p_3\) is the probability that Alice gets more heads than Bob, then by symmetry \(p_3 = p_2\) and \(p_1 + p_2 + p_3 = 1\). Therefore \(p_1 + 2p_2 = 1\). ie \(\frac12 p_1 + p_2 = \frac12\) therefore the answer is always \(\frac12\) for all values of \(n\).

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2016 Paper 1 Q13
D: 1500.0 B: 1500.0

An internet tester sends \(n\) e-mails simultaneously at time \(t=0\). Their arrival times at their destinations are independent random variables each having probability density function \(\lambda \e^{-\lambda t}\) (\(0\le t<\infty\), \( \lambda >0\)).

  1. The random variable \(T\) is the time of arrival of the e-mail that arrives first at its destination. Show that the probability density function of \(T\) is \[ n \lambda \e^{-n\lambda t}\,,\] and find the expected value of \(T\).
  2. Write down the probability that the second e-mail to arrive at its destination arrives later than time \(t\) and hence derive the density function for the time of arrival of the second e-mail. Show that the expected time of arrival of the second e-mail is \[ \frac{1}{\lambda} \left( \frac1{n-1} + \frac 1 n \right) \]

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(T > t) &= \mathbb{P}(\text{all emails slower than }t) \\ &&&= \left ( \int_t^{\infty} \lambda e^{-\lambda x} \d x \right)^n \\ &&&= \left ( [- e^{-\lambda x}]_t^\infty\right)^n\\ &&&= e^{-n\lambda t} \\ \Rightarrow && f_T(t) &= n \lambda e^{-n\lambda t} \\ \end{align*} Therefore \(T \sim \text{Exp}(n \lambda)\) and \(\E[T] = \frac{1}{n \lambda}\)
  2. Let \(T_2\) be the time until the second email arrives, then. \begin{align*} && \P(T_2 > t) &= \P(\text{all emails} > t) + \P(\text{all but 1 emails} > t) \\ &&&= e^{-n\lambda t} + n \cdot e^{-(n-1)\lambda t}(1-e^{-\lambda t}) \\ &&&= (1-n)e^{-n\lambda t} + n \cdot e^{-(n-1)\lambda t} \\ \Rightarrow && f_{T_2}(t) &= - \left ( (1-n) n \lambda e^{-n \lambda t} -n(n-1)\lambda e^{-(n-1)\lambda t} \right) \\ &&&= n(n-1) \lambda \left (e^{-(n-1)\lambda t} - e^{-n\lambda t} \right) \\ \Rightarrow && \E[T_2] &= \int_0^{\infty} t \cdot n(n-1) \lambda \left (e^{-(n-1)\lambda t} - e^{-n\lambda t} \right) \d t \\ &&&= \int_0^{\infty} \left (n \cdot t (n-1) \lambda e^{-(n-1)\lambda t} -(n-1)\cdot tn \lambda e^{-n\lambda t} \right) \d t \\ &&&= \frac{n}{\lambda(n-1)} - \frac{n-1}{\lambda n} \\ &&&= \frac{1}{\lambda} \left (1+\frac{1}{n-1}- \left (1 - \frac{1}{n} \right) \right) \\ &&&= \frac{1}{\lambda} \left ( \frac{1}{n-1} + \frac{1}{n} \right) \end{align*} (We can also view this second expectation as expected time for first email + expected time (of the remaining \(n-1\) emails) for the first email, and we can see that will have that form by the memorilessness property of exponentials)

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