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2007 Paper 1 Q11
D: 1500.0 B: 1484.0

A smooth, straight, narrow tube of length \(L\) is fixed at an angle of \(30^\circ\) to the horizontal. A~particle is fired up the tube, from the lower end, with initial velocity \(u\). When the particle reaches the upper end of the tube, it continues its motion until it returns to the same level as the lower end of the tube, having travelled a horizontal distance \(D\) after leaving the tube. Show that \(D\) satisfies the equation \[ 4gD^2 - 2 \sqrt{3} \left( u^2 - Lg \right)D - 3L \left( u^2 - gL \right) = 0 \] and hence that \[ \frac{{\rm d}D}{ {\rm d}L} = - \frac{ 2\sqrt{3}gD - 3(u^2-2gL)} { 8gD - 2 \sqrt{3} \left(u^2 - gL \right)}. \] The final horizontal displacement of the particle from the lower end of the tube is \(R\). Show that \(\dfrac{\d R}{\d L} = 0\) when \(2D = L \sqrt 3\), and determine, in terms of \(u\) and \(g\), the corresponding value of \(R\).

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2007 Paper 1 Q12
D: 1500.0 B: 1484.0

  1. A bag contains \(N\) sweets (where \(N \ge 2\)), of which \(a\) are red. Two sweets are drawn from the bag without replacement. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.
  2. There are two bags, each containing \(N\) sweets (where \(N \ge 2\)). The first bag contains \(a\) red sweets, and the second bag contains \(b\) red sweets. There is also a biased coin, showing Heads with probability \(p\) and Tails with probability \(q\), where \(p+q = 1\). The coin is tossed. If it shows Heads then a sweet is chosen from the first bag and transferred to the second bag; if it shows Tails then a sweet is chosen from the second bag and transferred to the first bag. The coin is then tossed a second time: if it shows Heads then a sweet is chosen from the first bag, and if it shows Tails then a sweet is chosen from the second bag. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.

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2007 Paper 1 Q13
D: 1500.0 B: 1469.5

A bag contains eleven small discs, which are identical except that six of the discs are blank and five of the discs are numbered, using the numbers 1, 2, 3, 4 and 5. The bag is shaken, and four discs are taken one at a time without replacement. Calculate the probability that:

  1. all four discs taken are numbered;
  2. all four discs taken are numbered, given that the disc numbered ``3'' is taken first;
  3. exactly two numbered discs are taken, given that the disc numbered ``3'' is taken first;
  4. exactly two numbered discs are taken, given that the disc numbered ``3'' is taken;
  5. exactly two numbered discs are taken, given that a numbered disc is taken first;
  6. exactly two numbered discs are taken, given that a numbered disc is taken.

Solution: There are many ways to do the counting in each question, possibly the clearest way is to always consider the order in which discs are taken, although all methods should work equally well. For some examples Bayes rule also offers a fast solution.

  1. There are we are choose \(4\) objects in order from \(5\) (ie \({^5\P_4}\)) to obtain valid draws, this is out of a total of picking \(4\) objects from \(11\) (\({^{11}\P_4}\)). Ie the probability is: \(\displaystyle \frac{{^5\P_4}}{{^{11}\P_4}} = \frac{5! \cdot 7!}{11!} = \frac{1}{66}\) Alternatively, there are \(\binom{5}{4}\) ways to choose four numbered discs, out of \(\binom{11}{4}\) ways to choose four discs. ie \(\displaystyle \binom{5}{4} \Big / \binom{11}{4} = \frac{5 \cdot 4! \cdot 7!}{11!} = \frac{5 \cdot 4 \cdot 3 \cdot 2}{11 \cdot 10 \cdot 9 \cdot 8} = \frac1{66}\)
  2. \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{^4\P_3 \big / {^{11}\P_4}}{1/11} \\ &= 11\cdot \frac{4!}{1!} \Bigg / \frac{11!}{7!} \\ &= \frac{4! \cdot 7! \cdot 11}{11!} \\ &= \frac{4\cdot 3 \cdot 2}{10 \cdot 9 \cdot 8} \\ &= \frac{1}{30} \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{3} \Bigg / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac{1}{30} \end{align*} Where we are calculating this as "choose one number", then "choose 3 more", which can happen ending up with 3, number, number, number in \(\binom{4}{3}\) ways, and there are \(11 \cdot \binom{10}{3}\) was overall. Another alternative using Bayes rule: \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \mathbb{P}( \text{first disc is 3} | \text{all four discs are numbered}) \frac{ \mathbb{P}( \text{all four discs are numbered} }{ \mathbb{P}( \text{first disc is 3} )} \\ &= \frac{\frac{1}{5} \cdot \frac{1}{66}}{\frac{1}{11}} \\ &= \frac1{30} \end{align*}
  3. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{3 \cdot {^4\P_1} \cdot {^{6}\P_2} \big / {^{11}\P_4}}{\frac1{11} } \\ &= \frac12 \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac12 \end{align*}
  4. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and 3 taken})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / \binom{11}{4}}{\frac{4}{11}} \\ &= \frac{\frac{2}{11}}{\frac4{11}} \\ &= \frac12 \end{align*} Using Bayes rule: \(\mathbb{P}( \text{3 taken}) = \frac{1}{11} + \frac{10}{11}\frac{1}{10} + \frac{10}{11}\frac{9}{10}\frac{1}{9} + \frac{10}{11}\frac{9}{10}\frac89\frac18 = \frac{4}{11}\) \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{3 taken | exactly two discs are numbered})\mathbb{P}(\text{exactly two discs are numbered})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\frac{4}{10} \cdot \binom{5}{2} \binom{6}{2} \Big / \binom{11}{4}}{4/11} \\ &= \frac{\frac4{10}{5 / 11}}{4/11} \\ &= \frac{1}{2} \end{align*}
  5. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc first}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc first})}{\mathbb{P}( \text{numbered disc first})} \\ &= \frac{3 \cdot {^5\P_1}\cdot{^4\P_1}\cdot{^6\P_2} \Big / {^{11}\P_4}}{\frac{5}{11}} \\ &= \frac{1}{2} \end{align*}
  6. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc taken})}{\mathbb{P}(\text{numbered disc taken})} \\ &= \frac{\mathbb{P}(\text{exactly two discs are numbered})}{1 - \mathbb{P}(\text{not numbered discs taken})} \\ &= \frac{\binom{5}{2}\binom{6}{2} \Big / \binom{11}{4}}{1 - \binom{6}{4} \Big / \binom{11}{4}} \\ &= \frac{\frac{5}{11}}{\frac{21}{22}} \\ &= \frac{10}{21} \neq \frac12 \end{align*}

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2007 Paper 1 Q14
D: 1500.0 B: 1484.0

The discrete random variable \(X\) has a Poisson distribution with mean \(\lambda\).

  1. Sketch the graph \(y=\l x+1 \r \e^{-x}\), stating the coordinates of the turning point and the points of intersection with the axes. It is known that \(\P(X \ge 2) = 1-p\), where \(p\) is a given number in the range \(0 < p <1\). Show that this information determines a unique value (which you should not attempt to find) of \(\lambda\).
  2. It is known (instead) that \(\P \l X = 1 \r = q\), where \(q\) is a given number in the range \(0 < q <1\). Show that this information determines a unique value of \(\lambda\) (which you should find) for exactly one value of \(q\) (which you should also find).
  3. It is known (instead) that \(\P \l X = 1 \, \vert \, X \le 2 \r = r\), where \(r\) is a given number in the range \(0

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