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2004 Paper 1 Q11
D: 1500.0 B: 1500.0

Two uniform ladders \(AB\) and \(BC\) of equal length are hinged smoothly at \(B\). The weight of \(AB\) is \(W\) and the weight of \(BC\) is \(4W \). The ladders stand on rough horizontal ground with \(\angle ABC=60^\circ\,\). The coefficient of friction between each ladder and the ground is \(\mu\). A decorator of weight \(7W\) begins to climb the ladder \(AB\) slowly. When she has climbed up \(\frac13\) of the ladder, one of the ladders slips. Which ladder slips, and what is the value of \(\mu\)?

Solution:

TikZ diagram
\begin{align*} \text{N2}(\rightarrow): && F_A - F_C &= 0\\ && F_A &= F_C \\ \text{N2}(\uparrow): && R_A + R_C - 7W - W - 4W &= 0\\ && R_A + R_C &= 12W \\ \overset{\curvearrowright}{A}: && \frac{1}{6}7W + \frac{1}{4}W + \frac{3}{4}4W - R_C &= 0 \\ \Rightarrow && R_C &= \frac{53}{12}W\\ \Rightarrow && R_A = 12W - \frac{53}{12}W &= \frac{91}{12}W \\ \overset{\curvearrowleft}{B}(AB): && \frac{1}{2}W + \frac{2}{3}7W - R_A+\sqrt{3}F_A &= 0 \\ \Rightarrow && F_A = \frac{1}{\sqrt{3}} \l \frac{91}{12}-\frac12-\frac{14}3\r W &= \frac{29}{12\sqrt{3}}W \end{align*} We know that the system is about to slip, so equality holds in one of \(F_A \leq \mu R_A\) or \(F_C \leq \mu R_C\). Since \(F_A = F_C\), we know it must occur for whichever of \(\mu R_A\) and \(\mu R_C\) is smaller. Since \(R_C\) is much smaller, this must be the ladder about to slip \(BC\) and \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]

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2004 Paper 1 Q12
D: 1500.0 B: 1529.3

In a certain factory, microchips are made by two machines. Machine A makes a proportion~\(\lambda\) of the chips, where \(0 < \lambda < 1\), and machine B makes the rest. A proportion \(p\) of the chips made by machine A are perfect, and a proportion \(q\) of those made by machine B are perfect, where \(0 < p < 1\) and \(0 < q < 1\). The chips are sorted into two groups: group 1 contains those that are perfect and group 2 contains those that are imperfect. In a large random sample taken from group 1, it is found that \(\frac 2 5\) were made by machine A. Show that \(\lambda\) can estimated as \[ {2q \over 3p + 2q}\;. \] Subsequently, it is discovered that the sorting process is faulty: there is a probability of \(\frac 14\) that a perfect chip is assigned to group 2 and a probability of \(\frac 14\) that an imperfect chip is assigned to group 1. Taking into account this additional information, obtain a new estimate of \(\lambda\,\).

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2004 Paper 1 Q13
D: 1500.0 B: 1458.1

  1. Three real numbers are drawn independently from the continuous rectangular distribution on \([ 0, 1 ]\,\). The random variable \(X\) is the maximum of the three numbers. Show that the probability that \(X \le 0.8\) is \(0.512\,\), and calculate the expectation of \(X\).
  2. \(N\) real numbers are drawn independently from a continuous rectangular distribution on \([ 0, a ]\,\). The random variable \(X\) is the maximum of the \(N\) numbers. A hypothesis test with a significance level of 5\% is carried out using the value, \(x\), of \(X \). The null hypothesis is that \(a=1\) and the alternative hypothesis is that \(a<1 \,\). The form of the test is such that \(H_0\) is rejected if \(x < c\,\), for some chosen number \(c\,\). Using the approximation \(2^{10} \approx 10^3\,\), determine the smallest integer value of \(N\) such that if \(x \le 0.8\) the null hypothesis will be rejected. With this value of \(N\), write down the probability that the null hypothesis is rejected if \(a = 0.8\,\), and find the probability that the null hypothesis is rejected if \(a = 0.9\,\).

Solution: \begin{align*} \P(X \leq 0.8) &= \P(X_1 \leq 0.8,X_2 \leq 0.8,X_3 \leq 0.8) \\ &= 0.8^3 \\ &= 0.512 \end{align*} \begin{align*} && \P(X < c) &= c^3 \\ \Rightarrow && f_X(x) &= 3x^2 \\ \Rightarrow && \E[X] &= \int_0^1 x \cdot (3x^2) \, dx \\ && &= \left [ \frac{3}{4}x^4 \right]_0^1 \\ &&&= \frac{3}{4} \end{align*} \(X\) is distributed the maximum of \(N\) numbers on \([0,a]\). \begin{align*} H_0 : & x= 1 \\ H_1 : & x < 1 \end{align*} \begin{align*} &&\P(X < c) &= c^N \\ &&&= \frac1{20} \\ \Rightarrow && N &= -\frac{\log(20)}{\log(c)} \end{align*} where \(c = 0.8\), we have \begin{align*} N &= \frac{\log(20)}{\log(5/4)} \\ &= \frac{\log(5)+\log(4)}{\log(5)-\log(4)} \\ &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} \end{align*} \begin{align*} && 2^{10} &\approx 10^{3} \\ && 10\log(2) &\approx 3 (\log(5) + \log(2)) \\ && 7\log(2) &\approx 3 \log(5) \\ && \frac{\log(5)}{2\log(2)} &\approx \frac{7}{6} \end{align*} \begin{align*} &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} &= \frac{\frac{7}{6} + 1}{\frac{7}{6} -1} \\ &= 13 \end{align*} Since \(2^{10} > 10^3\) then \(N=14\) is the value we seek. \(\P(X < 0.8 | a= 0.8) = 1\) \(\P(X < 0.8 | a= 0.9, N=14) = \frac{8^{14}}{9^{14}}\)

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2004 Paper 1 Q14
D: 1500.0 B: 1488.1

Three pirates are sharing out the contents of a treasure chest containing \(n\) gold coins and \(2\) lead coins. The first pirate takes out coins one at a time until he takes out one of the lead coins. The second pirate then takes out coins one at a time until she draws the second lead coin. The third pirate takes out all the gold coins remaining in the chest. Find:

  1. the probability that the first pirate will have some gold coins;
  2. the probability that the second pirate will have some gold coins;
  3. the probability that all three pirates will have some gold coins.

Solution:

  1. The first pirate will have some gold coins as long as the very first coin drawn is a gold coin, ie \(\frac{n}{n+2}\).
  2. The probability the second pirate will have some gold coins is the probability the two lead coins are separate. There are \((n+2)!\) ways to arrange the coins, and there are \((n+1)! \cdot 2\) ways to arrange the coins where they are together, therefore the probability is: \[ 1 - \frac{2(n+1)!}{(n+2)!} = 1 - \frac{2}{n+2} = \frac{n}{n+2} \]
  3. For all three pirates to have some gold coins we need the lead coins to be separate and not first or last. If we line up all \(n\) gold coins, there are \(n-1\) gaps between them we could place the \(2\) lead coins in, therefore \(\binom{n-1}{2}\) ways to place the lead coins with the restriction. Without the restriction there are \(\binom{n+2}{2}\) ways to choose where to put the coins, therefore \begin{align*} && P &= \frac{\binom{n-1}{2}}{\binom{n+2}{2}} \\ &&&= \frac{(n-1)(n-2)}{(n+2)(n+1)} \end{align*} [Notice this is clearly \(0\) if there are only \(1\) or \(2\) gold coins]

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