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2016 Paper 2 Q11
D: 1600.0 B: 1484.0

  1. Two particles move on a smooth horizontal surface. The positions, in Cartesian coordinates, of the particles at time \(t\) are \((a+ut\cos\alpha \,,\, ut\sin\alpha)\) and \((vt\cos\beta\,,\, b+vt\sin\beta )\), where \(a\), \(b\), \(u\) and \(v\) are positive constants, \(\alpha\) and \(\beta\) are constant acute angles, and \(t\ge0\). Given that the two particles collide, show that \[ u \sin(\theta+\alpha) = v\sin(\theta +\beta)\,, \] where \(\theta \) is the acute angle satisfying \(\tan\theta = \dfrac b a\).
  2. A gun is placed on the top of a vertical tower of height \(b\) which stands on horizontal ground. The gun fires a bullet with speed \(v\) and (acute) angle of elevation \(\beta\). Simultaneously, a target is projected from a point on the ground a horizontal distance \(a\) from the foot of the tower. The target is projected with speed \(u\) and (acute) angle of elevation \(\alpha\), in a direction directly away from the tower. Given that the target is hit before it reaches the ground, show that \[ 2u\sin\alpha (u\sin\alpha - v\sin\beta)>bg\,. \] Explain, with reference to part (i), why the target can only be hit if \(\alpha>\beta\).

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2016 Paper 2 Q12
D: 1600.0 B: 1503.2

Starting with the result \(\P(A\cup B) = \P(A)+P(B) - \P(A\cap B)\), prove that \[ \P(A\cup B\cup C) = \P(A)+\P(B)+\P(C) - \P(A\cap B) - \P(B\cap C) - \P(C \cap A) + \P(A\cap B\cap C) \,. \] Write down, without proof, the corresponding result for four events \(A\), \(B\), \(C\) and \(D\). A pack of \(n\) cards, numbered \(1, 2, \ldots, n\), is shuffled and laid out in a row. The result of the shuffle is that each card is equally likely to be in any position in the row. Let \(E_i\) be the event that the card bearing the number \(i\) is in the \(i\)th position in the row. Write down the following probabilities:

  1. \(\P(E_i)\);
  2. \(\P(E_i\cap E_j)\), where \(i\ne j\);
  3. \(\P(E_i\cap E_j\cap E_k)\), where \(i\ne j\), \(j\ne k\) and \(k\ne i\).
Hence show that the probability that at least one card is in the same position as the number it bears is \[ 1 - \frac 1 {2!} + \frac 1{3!} - \cdots + (-1)^{n+1} \frac 1 {n!}\,. \] Find the probability that exactly one card is in the same position as the number it bears

Solution: \begin{align*} && \mathbb{P}(A \cup B \cup C) &= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cup B) \cap C) \tag{applying with \(A\cup B\) and \(C\)} \\ &&&= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \tag{applying with \(A\) and \(B\)}\\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \left ( \mathbb{P}(A \cap C) +\mathbb{P}(B \cap C) - \mathbb{P}( (A \cap C) \cap (B \cap C) )\right) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) +\mathbb{P}(C)- \mathbb{P}(A\cap B)- \mathbb{P}(A \cap C) -\mathbb{P}(B \cap C)+\mathbb{P}( A \cap B \cap C) \end{align*} \[ \mathbb{P}(A_1 \cup A_2 \cup A_3 \cup A_4) = \sum_i \mathbb{P}(A_i) - \sum_{i \neq j} \mathbb{P}(A_i \cap A_j) + \sum_{i \neq j \neq j} \mathbb{P}(A_i \cap A_j \cap A_k) - \mathbb{P}(A_1 \cap A_2 \cap A_3 \cap A_4) \]

  1. \(\mathbb{P}(E_i) = \frac{1}{n}\)
  2. \(\mathbb{P}(E_i \cap E_j) = \frac{1}{n} \cdot \frac{1}{n-1} = \frac{1}{n(n-1)}\)
  3. \(\mathbb{P})(E_i \cap E_j \cap E_k) = \frac{1}{n(n-1)(n-2)}\)
First notice that the probability that \(k\) (or more) cards are in the correct place is \(\frac{(n-k)!}{n!}\) (place the other \(n-k\) cards in any order. We are interested in: \begin{align*} \mathbb{P} \left ( \bigcup_{i=1}^n E_i \right) &= \sum_{i} \mathbb{P}(E_i) - \sum_{i \neq j} \mathbb{P}(E_i \cap E_j) + \sum_{i \neq j \neq k} \mathbb{P}(E_i \cap E_j \cap E_k) - \cdots \\ &= \sum_i \frac1n - \sum_{i\neq j} \frac{1}{n(n-1)} + \sum_{i \neq j \neq k} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \sum_{i_1 \neq i_2 \neq \cdots \neq i_k} \frac{(n-k)!}{n!} + \cdots\\ &= 1 - \binom{n}{2} \frac{1}{n(n-1)} + \binom{n}{3} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \binom{n}{k} \frac{(n-k)}{n!} + \cdots \\ &= 1 - \frac12 + \frac1{3!} - \cdots + (-1)^{k+1} \frac{n!}{k!(n-k)!} \frac{(n-k)!}{n!} + \cdots \\ &= 1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n+1} \frac{1}{n!} \end{align*} The probability exactly one card is in the right place is the probability none of the other \(n-1\) are in the right place, which is: \(\frac1n \left (1 - \left (1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n} \frac{1}{(n-1)!} \right) \right)\) but there are also \(n\) cards we can choose to be the card in the right place, hence \(\frac{1}{2!} - \frac{1}{3!} + \cdots +(-1)^n \frac{1}{(n-1)!}\)

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2016 Paper 2 Q13
D: 1600.0 B: 1516.0

  1. The random variable \(X\) has a binomial distribution with parameters \(n\) and \(p\), where \(n=16\) and \(p=\frac12\). Show, using an approximation in terms of the standard normal density function $\displaystyle \tfrac{1}{\sqrt{2\pi}} \, \e ^{-\frac12 x^2} $, that \[ \P(X=8) \approx \frac 1{2\sqrt{2\pi}} \,. \]
  2. By considering a binomial distribution with parameters \(2n\) and \(\frac12\), show that \[ (2n)! \approx \frac {2^{2n} (n!)^2}{\sqrt{n\pi}} \,. \]
  3. By considering a Poisson distribution with parameter \(n\), show that \[ n! \approx \sqrt{2\pi n\, } \, \e^{-n} \, n^n \,. \]

Solution:

  1. \(X \sim B(16, \tfrac12)\), then \(X \approx N(8, 2^2)\), in particular \begin{align*} && \mathbb{P}(X = 8) &\approx \mathbb{P} \left ( 8 - \frac12 \leq 2Z + 8 \leq 8 + \frac12 \right) \\ &&&= \mathbb{P} \left (-\frac14 \leq Z \leq \frac14 \right) \\ &&&= \int_{-\frac14}^{\frac14} \frac{1}{\sqrt{2 \pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2\pi}} \int_{-\frac14}^{\frac14} 1\d x\\ &&&= \frac{1}{2 \sqrt{2\pi}} \end{align*}
  2. Suppose \(X \sim B(2n, \frac12)\) then \(X \approx N(n, \frac{n}{2})\), and \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left ( n - \frac12 \leq \sqrt{\frac{n}{2}} Z + n \leq n + \frac12 \right) \\ &&&= \mathbb{P} \left ( - \frac1{\sqrt{2n}} \leq Z \leq \frac1{\sqrt{2n}}\right) \\ &&&= \int_{-\frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2 \pi}} e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{n\pi}}\\ \Rightarrow && \binom{2n}{n}\frac1{2^n} \frac{1}{2^n} & \approx \frac{1}{\sqrt{n \pi}} \\ \Rightarrow && (2n)! &\approx \frac{2^{2n}(n!)^2}{\sqrt{n\pi}} \end{align*}
  3. \(X \sim Po(n)\), then \(X \approx N(n, (\sqrt{n})^2)\), therefore \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left (-\frac12 \leq \sqrt{n} Z \leq \frac12 \right) \\ &&&= \int_{-\frac{1}{2 \sqrt{n}}}^{\frac{1}{2 \sqrt{n}}} \frac{1}{\sqrt{2\pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && e^{-n} \frac{n^n}{n!} & \approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && n! &\approx \sqrt{2 \pi n} e^{-n}n^n \end{align*}

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