Continuous Probability Distributions and Random Variables

Showing 1-18 of 18 problems
2024 Paper 3 Q12
D: 1500.0 B: 1500.0

  1. A point is chosen at random in the square \(0 \leqslant x \leqslant 1\), \(0 \leqslant y \leqslant 1\), so that the probability that a point lies in any region is equal to the area of that region. \(R\) is the random variable giving the distance of the point from the origin. Show that the cumulative distribution function of \(R\) is given by \[\mathrm{P}(R \leqslant r) = \sqrt{r^2 - 1} + \tfrac{1}{4}\pi r^2 - r^2 \cos^{-1}(r^{-1}),\] when \(1 \leqslant r \leqslant \sqrt{2}\). What is the cumulative distribution function when \(0 \leqslant r \leqslant 1\)?
  2. Show that \(\displaystyle\mathrm{E}(R) = \frac{2}{3}\int_1^{\sqrt{2}} \frac{r^2}{\sqrt{r^2-1}}\,\mathrm{d}r\).
  3. Show further that \(\mathrm{E}(R) = \frac{1}{3}\Bigl(\sqrt{2} + \ln\bigl(\sqrt{2}+1\bigr)\Bigr)\).

2023 Paper 2 Q12
D: 1500.0 B: 1500.0

Each of the independent random variables \(X_1, X_2, \ldots, X_n\) has the probability density function \(\mathrm{f}(x) = \frac{1}{2}\sin x\) for \(0 \leqslant x \leqslant \pi\) (and zero otherwise). Let \(Y\) be the random variable whose value is the maximum of the values of \(X_1, X_2, \ldots, X_n\).

  1. Explain why \(\mathrm{P}(Y \leqslant t) = \big[\mathrm{P}(X_1 \leqslant t)\big]^n\) and hence, or otherwise, find the probability density function of \(Y\).
Let \(m(n)\) be the median of \(Y\) and \(\mu(n)\) be the mean of \(Y\).
  1. Find an expression for \(m(n)\) in terms of \(n\). How does \(m(n)\) change as \(n\) increases?
  2. Show that \[\mu(n) = \pi - \frac{1}{2^n}\int_0^{\pi} (1-\cos x)^n\,\mathrm{d}x\,.\]
    1. Show that \(\mu(n)\) increases with \(n\).
    2. Show that \(\mu(2) < m(2)\).

2022 Paper 2 Q12
D: 1500.0 B: 1500.0

The random variable \(X\) has probability density function \[\mathrm{f}(x) = \begin{cases} kx^n(1-x) & 0 \leqslant x \leqslant 1\,,\\ 0 & \text{otherwise}\,,\end{cases}\] where \(n\) is an integer greater than 1.

  1. Show that \(k = (n+1)(n+2)\) and find \(\mu\), where \(\mu = \mathrm{E}(X)\).
  2. Show that \(\mu\) is less than the median of \(X\) if \[6 - \frac{8}{n+3} < \left(1 + \frac{2}{n+1}\right)^{n+1}.\] By considering the first four terms of the expansion of the right-hand side of this inequality, or otherwise, show that the median of \(X\) is greater than \(\mu\).
  3. You are given that, for positive \(x\), \(\left(1 + \dfrac{1}{x}\right)^{x+1}\) is a decreasing function of \(x\). Show that the mode of \(X\) is greater than its median.

Show Solution
  1. \(\,\) \begin{align*} && 1 &= \int_0^1 kx^n(1-x) \d x \\ &&&= k\left [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^1 \\ &&&= \frac{k}{(n+1)(n+2)} \\ \Rightarrow && k &= (n+1)(n+2) \\ \\ && \E[X] &= \int_0^1 kx^{n+1}(1-x) \d x \\ &&&= k\left [ \frac{x^{n+2}}{n+2} - \frac{x^{n+3}}{n+3} \right]_0^1 \\ &&&= \frac{(n+1)(n+2)}{(n+2)(n+3)} \\ &&&= \frac{n+1}{n+3} \end{align*}
  2. If \(\mu\) is less than the median then \begin{align*} && \frac12 &> \int_0^{\mu} kx^{n}(1-x) \d x \\ &&&= \left [k \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^\mu \\ &&&= \mu^{n+1}\left ((n+2) - (n+1) \frac{n+1}{n+3} \right) \\ \Rightarrow && \left ( 1 + \frac{2}{n+1} \right)^{n+1} &> 2 \frac{(n+2)(n+3) - (n+1)^2}{n+3} \\ &&&= \frac{6n+10}{n+3} = 6 - \frac{8}{n+3} \end{align*} \begin{align*} && \left ( 1 + \frac{2}{n+1} \right)^{n+1} &= 1 + (n+1) \frac{2}{n+1} + \frac{(n+1)n}{2} \frac{4}{(n+1)^2} + \frac{(n+1)n(n-1)}{6} \frac{8}{(n+1)^3} + \cdots \\ &&&= 1 + 2 + \frac{2n}{n+1} + \frac{4n(n-1)}{3(n+1)^2} + \cdots \\ &&&= 5 - \frac{2}{n+1} + \frac{4((n+1)^2-3(n+1)+3)}{3(n+1)^2} \\ &&&= 6 + \frac13 - \frac{6}{n+1} - \frac{4}{3(n+1)^2} \\ &&&> 6 - \frac{8}{n+3} \end{align*} as required.
  3. To find the mode, we need to find the maximum of \(x^n(1-x)\) which occurs when \(nx^{n-1}(1-x) - x^n = x^{n-1}(n-(n+1)x)\) ie where \(x = \frac{n}{n+1}\), therefore we need to look at: \begin{align*} && \mathbb{P}(X \leq \tfrac{n}{n+1}) &= \int_0^{n/(n+1)} k x^n(1-x) \d x \\ &&&= \frac{n^{n+1}}{(n+1)^{n+1}} \left ( (n+2) - (n+1) \frac{n}{n+1} \right) \\ &&&= 2\frac{n^{n+1}}{(n+1)^{n+1}} \\ &&&= 2 \left ( \frac{1}{1+\frac1n} \right)^{n+1} \\ &&&\geq 2 \frac{1}{(1 + \frac11)^2} = \frac12 \end{align*} as required
2022 Paper 3 Q12
D: 1500.0 B: 1500.0

  1. The point \(A\) lies on the circumference of a circle of radius \(a\) and centre \(O\). The point \(B\) is chosen at random on the circumference, so that the angle \(AOB\) has a uniform distribution on \([0, 2\pi]\). Find the expected length of the chord \(AB\).
  2. The point \(C\) is chosen at random in the interior of a circle of radius \(a\) and centre \(O\), so that the probability that it lies in any given region is proportional to the area of the region. The random variable \(R\) is defined as the distance between \(C\) and \(O\). Find the probability density function of \(R\). Obtain a formula in terms of \(a\), \(R\) and \(t\) for the length of a chord through \(C\) that makes an acute angle of \(t\) with \(OC\). Show that as \(C\) varies (with \(t\) fixed), the expected length \(\mathrm{L}(t)\) of such chords is given by \[ \mathrm{L}(t) = \frac{4a(1-\cos^3 t)}{3\sin^2 t}\,. \] Show further that \[ \mathrm{L}(t) = \frac{4a}{3}\left(\cos t + \tfrac{1}{2}\sec^2(\tfrac{1}{2}t)\right). \]
  3. The random variable \(T\) is uniformly distributed on \([0, \frac{1}{2}\pi]\). Find the expected value of \(\mathrm{L}(T)\).

2021 Paper 3 Q11
D: 1500.0 B: 1500.0

The continuous random variable \(X\) has probability density function \[ f(x) = \begin{cases} \lambda e^{-\lambda x} & \text{for } x \geqslant 0, \\ 0 & \text{otherwise,} \end{cases} \] where \(\lambda\) is a positive constant. The random variable \(Y\) is the greatest integer less than or equal to \(X\), and \(Z = X - Y\).

  1. Show that, for any non-negative integer \(n\), \[ \mathrm{P}(Y = n) = (1 - e^{-\lambda})\,e^{-n\lambda}. \]
  2. Show that \[ \mathrm{P}(Z < z) = \frac{1 - e^{-\lambda z}}{1 - e^{-\lambda}} \qquad \text{for } 0 \leqslant z \leqslant 1. \]
  3. Evaluate \(\mathrm{E}(Z)\).
  4. Obtain an expression for \[ \mathrm{P}(Y = n \text{ and } z_1 < Z < z_2), \] where \(0 \leqslant z_1 < z_2 \leqslant 1\) and \(n\) is a non-negative integer. Determine whether \(Y\) and \(Z\) are independent.

Show Solution
  1. \(\,\) \begin{align*} && \mathbb{P}(Y = n) &= \mathbb{P}(X \in [n, n+1)) \\ &&&= \int_n^{n+1} \lambda e^{-\lambda x} \d x \\ &&&= \left [-e^{-\lambda x} \right]_n^{n+1} \\ &&&= e^{-\lambda n} - e^{-\lambda(n+1)} \\ &&&= e^{-\lambda n}(1- e^{-\lambda}) \end{align*}
  2. \(,\) \begin{align*} && \mathbb{P}(Z < z) &= \sum_{i=0}^{\infty} \mathbb{P}(X \in (n, n+z)) \\ &&&= \sum_{i=0}^{\infty} \int_{n}^{n+z} \lambda e^{-\lambda x} \d x \\ &&&= \sum_{i=0}^{\infty} [-e^{-\lambda x}]_{n}^{n+z} \\ &&&= \sum_{i=0}^{\infty} (1-e^{-\lambda x})e^{-\lambda n} \\ &&&= \frac{1-e^{-\lambda x}}{1-e^{-\lambda}} \end{align*}
  3. Give the cdf of \(Z\), we see that \(f_Z(z) = \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}}\) so \begin{align*} && \E[Z] &= \int_0^1 z \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \int_0^1 ze^{-\lambda z} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( \left [-\frac{1}{\lambda} ze^{-\lambda z} \right]_0^1+\int_0^1 \frac{1}{\lambda} e^{-\lambda z} \d z \right) \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( -\frac{e^{-\lambda}}{\lambda} + \frac{1-e^{-\lambda}}{\lambda^2} \right) \\ &&&= \frac{1-e^{-\lambda}(1+\lambda)}{\lambda (1-e^{-\lambda})} \end{align*}
  4. \(\,\) \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2)&= \mathbb{P}(X \in (n+z_1, n+z_2) ) \\ &&&= \int_{n+z_1}^{n+z_2} \lambda e^{-\lambda x} \d x \\ &&&= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \end{align*} Note that \(\mathbb{P}(z_1 < Z < z_2) = \mathbb{P}( Z < z_2) -\mathbb{P}(Z< z_1) =\frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}}\) Therefore \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2) &= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \\ &&&= e^{-\lambda n}(1-e^{-\lambda}) \frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}} \\ &&&= \mathbb{P}(Y=n) \mathbb{P}(z_1 < Z < z_2) \end{align*} So they are independent, which is to be expected from the memorylessness property of the exponential distribution.
2020 Paper 3 Q11
D: 1500.0 B: 1500.0

The continuous random variable \(X\) is uniformly distributed on \([a,b]\) where \(0 < a < b\).

  1. Let \(\mathrm{f}\) be a function defined for all \(x \in [a,b]\)
    • with \(\mathrm{f}(a) = b\) and \(\mathrm{f}(b) = a\),
    • which is strictly decreasing on \([a,b]\),
    • for which \(\mathrm{f}(x) = \mathrm{f}^{-1}(x)\) for all \(x \in [a,b]\).
    The random variable \(Y\) is defined by \(Y = \mathrm{f}(X)\). Show that \[ \mathrm{P}(Y \leqslant y) = \frac{b - \mathrm{f}(y)}{b - a} \quad \text{for } y \in [a,b]. \] Find the probability density function for \(Y\) and hence show that \[ \mathrm{E}(Y^2) = -ab + \int_a^b \frac{2x\,\mathrm{f}(x)}{b-a} \; \mathrm{d}x. \]
  2. The random variable \(Z\) is defined by \(\dfrac{1}{Z} + \dfrac{1}{X} = \dfrac{1}{c}\) where \(\dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}\). By finding the variance of \(Z\), show that \[ \ln\left(\frac{b-c}{a-c}\right) < \frac{b-a}{c}. \]

2019 Paper 2 Q12
D: 1500.0 B: 1500.0

The random variable \(X\) has the probability density function on the interval \([0, 1]\): $$f(x) = \begin{cases} nx^{n-1} & 0 \leq x \leq 1, \\ 0 & \text{elsewhere}, \end{cases}$$ where \(n\) is an integer greater than 1.

  1. Let \(\mu = E(X)\). Find an expression for \(\mu\) in terms of \(n\), and show that the variance, \(\sigma^2\), of \(X\) is given by $$\sigma^2 = \frac{n}{(n + 1)^2(n + 2)}.$$
  2. In the case \(n = 2\), show without using decimal approximations that the interquartile range is less than \(2\sigma\).
  3. Write down the first three terms and the \((k + 1)\)th term (where \(0 \leq k \leq n\)) of the binomial expansion of \((1 + x)^n\) in ascending powers of \(x\). By setting \(x = \frac{1}{n}\), show that \(\mu\) is less than the median and greater than the lower quartile. Note: You may assume that $$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots < 4.$$

Show Solution
  1. \(\,\) \begin{align*} && \mu &= \E[X] \\ &&&= \int_0^1 x f(x) \d x \\ &&&= \int_0^1 nx^n \d x \\ &&&= \frac{n}{n+1} \\ \\ && \var[X] &= \sigma^2 \\ &&&= \E[X^2] - \mu^2 \\ &&&= \int_0^1 x^2 f(x) \d x - \mu^2 \\ &&&= \int_0^1 nx^{n+1} \d x - \mu^2 \\ &&&= \frac{n}{n+2} - \frac{n^2}{(n+1)^2} \\ &&&= \frac{n(n+1)^2 - n^2(n+2)}{(n+1)^2(n+2)} \\ &&&= \frac{n}{(n+1)^2(n+2)} \end{align*}
  2. \(\,\) \begin{align*} && \frac14 &= \int_0^{Q_1} 2x \d x \\ &&&= Q_1^2 \\ \Rightarrow && Q_1 &= \frac12 \\ && \frac34 &= \int_0^{Q_3} 2x \d x \\ &&&= Q_3^2 \\ \Rightarrow && Q_3 &= \frac{\sqrt{3}}2 \\ \\ \Rightarrow && IQR &= Q_3 - Q_1 = \frac{\sqrt{3}-1}{2} \\ && 2 \sigma &= 2\sqrt{\frac{2}{3^2 \cdot 4}} \\ &&&= \frac{\sqrt{2}}{3} \\ \\ && 2\sigma - IRQ &= \frac{\sqrt{2}}{3} - \frac{\sqrt{3}-1}{2} \\ &&&= \frac{2\sqrt{2}-3\sqrt{3}+3}{6} \\ && (3+2\sqrt{2})^2 &= 17+12\sqrt{2} > 29 \\ && (3\sqrt{3})^2 &= 27 \end{align*} Therefore \(2\sigma > IQR\)
  3. \[ (1+x)^n = 1 + nx + \frac{n(n-1)}2 x^2 + \cdots + \binom{n}{k} x^k+ \cdots \] \begin{align*} && Q_1^{-n} &= 4 \\ && Q_2^{-n} &= 2\\ && \mu &=\frac{n}{n+1} \\ \Rightarrow && \mu^{-n} &= \left (1 + \frac1n \right)^n\\ &&&\geq 1 + n \frac1n + \cdots > 2 \\ \Rightarrow && \mu &< Q_2 \\ \\ && \mu^{-n} &= \left (1 + \frac1n \right)^n\\ &&&= 1 + n \frac1n + \frac{n(n-1)}{2!} \frac{1}{n^2} + \cdots + \frac{n(n-1) \cdots (n-k+1)}{k!} \frac{1}{n^k} + \cdots \\ &&&= 1 + 1 + \left (1 - \frac1n \right ) \frac1{2!} + \cdots + \left (1 - \frac1n \right)\cdot\left (1 - \frac2n \right) \cdots \left (1 - \frac{k-1}n \right) \frac{1}{k!} + \cdots \\ &&&< 1 + 1 + \frac1{2!} + \cdots + \frac1{k!} \\ &&&< 4 \\ \Rightarrow && \mu &> Q_1 \end{align*}
2014 Paper 1 Q13
D: 1500.0 B: 1483.3

A continuous random variable \(X\) has a triangular distribution, which means that it has a probability density function of the form \[ \f(x) = \begin{cases} \g(x) & \text{for \(a< x \le c\)} \\ \h(x) & \text{for \(c\le x < b\)} \\ 0 & \text{otherwise,} \end{cases} \] where \(\g(x)\) is an increasing linear function with \(\g(a)=0\), \(\h(x)\) is a decreasing linear function with \(\h(b) =0\), and \(\g(c)=\h(c)\). Show that \(\g(x) = \dfrac{2(x-a)}{(b-a)(c-a)}\) and find a similar expression for \(\h(x)\).

  1. Show that the mean of the distribution is \(\frac13(a+b+c)\).
  2. Find the median of the distribution in the different cases that arise.

Show Solution
Since \(\int f(x) \, dx = 1\), and \(f(x)\) is a triangle with base \(b-a\), it must have height \(\frac{2}{b-a}\) in order to have the desired area. Since \(g(a) = 0, g(c) = \frac{2}{b-a}\), \(g(x) = A(x-a)\) and \(\frac{2}{b-a} = A (c-a) \Rightarrow g(x) = \frac{2(x-a)}{(b-a)(c-a)}\) as required. Similarly, \(h(x) = B(x-b)\) and \(\frac{2}{b-a} = B(c-b) \Rightarrow h(x) = \frac{2(b-x)}{(b-a)(b-c)}\) The mean of the distribution will be: \begin{align*} \int_a^b xf(x) \, dx &= \int_a^c xg(x) \, dx + \int_c^b xh(x) \, dx \\ &= \frac{2}{(b-a)(c-a)} \int_a^c x(x-a) dx + \frac{2}{(b-a)(b-c)} \int_c^b x(b-x) \, dx \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \left [ \frac{x^3}{3} - a\frac{x^2}{2} \right ]_a^c + \frac{1}{b-c} \left [ b\frac{x^2}{2} - \frac{x^3}{3} \right ]_c^b\r \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \l \frac{c^3}{3} - a\frac{c^2}{2} - \frac{a^3}{3} + \frac{a^3}{2} \r + \frac{1}{b-c} \l \frac{b^3}{2} - \frac{b^3}{3} - \frac{bc^2}{2} + \frac{c^3}{3} \r \r \\ &= \frac{2}{(b-a)} \l \l \frac{c^2+ac+a^2}{3} - \frac{a(a+c)}{2} \r +\l \frac{b(b+c)}{2} - \frac{b^2+bc+c^2}{3} \r\r \\ &= \frac{2}{(b-a)} \l \frac{2c^2+2ac+2a^2}{6} - \frac{3a^2+3ac}{6} + \frac{3b^2+3bc}{6} - \frac{2b^2+2bc+2c^2}{6} \r \\ &= \frac{2}{(b-a)} \l \frac{-a^2+b^2-ac+bc}{6} \r \\ &= \frac{a+b+c}{3} \\ \end{align*} The median \(M\) satisfies: \begin{align*} && \int_a^M f(x) \, dx &= \frac12 \\ \end{align*} The left hand triangle will have area: \(\frac{c-a}{b-a}\) which will be \(\geq \frac12\) if \(c \geq \frac{a+b}{2}\). In this case we need \begin{align*} && \frac{(M-a)^2}{(b-a)(c-a)} &= \frac12 \\ \Rightarrow && M &= a + \sqrt{\frac12 (b-a)(c-a)} \end{align*} Otherwise, we need: \begin{align*} && \frac{(b-M)^2}{(b-a)(b-c)} &= \frac12 \\ \Rightarrow && M &= b - \sqrt{\frac12 (b-a)(b-c)} \end{align*} These are consistent, if \(c = \frac{b+a}{2}\)
2012 Paper 1 Q12
D: 1484.0 B: 1516.0

Fire extinguishers may become faulty at any time after manufacture and are tested annually on the anniversary of manufacture. The time \(T\) years after manufacture until a fire extinguisher becomes faulty is modelled by the continuous probability density function \[ f(t) = \begin{cases} \frac{2t}{(1+t^2)^2}& \text{for \(t\ge0\)}\,,\\[4mm] \ \ \ \ 0& \text{otherwise}. \end{cases} \] A faulty fire extinguisher will fail an annual test with probability \(p\), in which case it is destroyed immediately. A non-faulty fire extinguisher will always pass the test. All of the annual tests are independent. Show that the probability that a randomly chosen fire extinguisher will be destroyed exactly three years after its manufacture is \(p(5p^2-13p +9)/10\). Find the probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture.

Show Solution
The probability it becomes faulty in each year is: \begin{align*} \mathbb{P}(\text{faulty in Y}1) &= \int_0^1 \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_0^1 \\ &= 1 - \frac{1}{2} = \frac{1}{2} \\ \mathbb{P}(\text{faulty in Y}2) &= \frac{1}{2} - \frac{1}{5} = \frac{3}{10} \\ \mathbb{P}(\text{faulty in Y}3) &= \frac{1}{5} - \frac{1}{10} = \frac{1}{10} \end{align*} The probability of failing for the first time after exactly \(3\) years is: \begin{align*} \mathbb{P}(\text{faulty in Y1, }PPF) &+ \mathbb{P}(\text{faulty in Y2, }PF) + + \mathbb{P}(\text{faulty in Y3, }F) \\ &= \frac12 (1-p)^2p + \frac3{10}(1-p)p + \frac1{10}p \\ &= \frac{p}{10} \l 5(1-p)^2 + 3(1-p) + 1 \r \\ &= \frac{p}{10} \l 5 - 10p + 5p^2 + 3 -3p +1 \r \\ &= \frac{p}{10} \l 9 - 13p + 5p^2 \r \end{align*} as required. The probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture is: \begin{align*} \mathbb{P}(\text{faulty 18 months after} | \text{fails after 3 tries}) &= \frac{\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})}{\mathbb{P}(\text{fails after exactly 3 tries})} \end{align*} We can compute \(\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})\) by looking at \(2\) cases, fails between \(12\) months and \(18\) years, and between \(0\) years and \(1\) year. \begin{align*} \mathbb{P}(\text{faulty between 1y and 18m}) &= \int_{1}^{\frac32} \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_{1}^{\frac32} \\ &= \frac12 - \frac{4}{13} = \frac{5}{26} \\ \end{align*} So the probability is: \begin{align*} \mathbb{P} &= \frac{\frac{5}{26}(1-p)p + \frac12(1-p)^2p}{\frac{p}{10} \l 9 - 13p + 5p^2 \r} \\ &= \frac{\frac{25}{13}(1-p) + 5(1-p)^2}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 5 + 13(1-p) \r}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 18 - 13p \r}{9 - 13p + 5p^2} \\ \end{align*}
2011 Paper 1 Q13
D: 1484.0 B: 1471.5

In this question, you may use without proof the following result: \[ \int \sqrt{4-x^2}\, \d x = 2 \arcsin (\tfrac12 x ) + \tfrac 12 x \sqrt{4-x^2} +c\,. \] A random variable \(X\) has probability density function \(\f\) given by \[ \f(x) = \begin{cases} 2k & -a\le x <0 \\[3mm] k\sqrt{4-x^2} & \phantom{-} 0\le x \le 2 \\[3mm] 0 & \phantom{-}\text{otherwise}, \end{cases} \] where \(k\) and \(a\) are positive constants.

  1. Find, in terms of \(a\), the mean of \(X\).
  2. Let \(d\) be the value of \(X\) such that \(\P(X> d)=\frac1 {10}\,\). Show that \(d < 0\) if \(2a> 9\pi\) and find an expression for \(d\) in terms of \(a\) in this case.
  3. Given that \(d=\sqrt 2\), find \(a\).

Show Solution
First notice that \begin{align*} && 1 &= \int_{-a}^2 f(x) \d x \\ &&&= 2ka + k\pi \\ \Rightarrow && k &= (\pi + 2a)^{-1} \end{align*}
  1. \(\,\) \begin{align*} && \E[X] &= \int_{-a}^2 x f(x) \d x\\ &&&= \int_{-a}^0 2kx \d x + k\int_0^{2} x\sqrt{4-x^2} \d x\\ &&&= \left [kx^2 \right]_{-a}^0 +k \left [-\frac13(4-x^2)^{\frac32} \right]_0^2 \\ &&&= -ka^2 + \frac83k \\ &&&= \frac{\frac83-a^2}{\pi + 2a} \end{align*}
  2. Consider \(\mathbb{P}(X < 0)\) then \(d < 0 \Leftrightarrow \mathbb{P}(X < 0) > \frac{9}{10}\) \begin{align*} && \frac{9}{10} &< \mathbb{P}(X < 0) \\ &&&= \int_{-a}^0 2k \d x \\ &&&= \frac{2a}{\pi+2a} \\ \Leftrightarrow && 9\pi &< 2a \\ \\ && \frac{9}{10} &= \int_{-a}^d 2k \d x \\ &&&= \frac{2(d+a)}{\pi + 2a} \\ \Rightarrow && 9\pi &= 2a + 20d \\ \Rightarrow && d &= \frac{2a-9\pi}{20} \end{align*}
  3. Suppose \(d=\sqrt 2\) then \begin{align*} && \frac1{10} &= \int_{\sqrt{2}}^2 f(x) \d x \\ &&&= \int_{\sqrt{2}}^2 k\sqrt{4-x^2} \d x \\ &&&= k\left [ 2 \sin^{-1} \tfrac12 x + \tfrac12 x \sqrt{4-x^2}\right]_{\sqrt{2}}^2 \\ &&&= k\left (\pi -\frac{\pi}{2} - 1 \right) \\ \Rightarrow && \pi + 2a &= 5\pi - 10 \\ \Rightarrow && a &= 2\pi-5 \end{align*}
2006 Paper 2 Q14
D: 1600.0 B: 1500.0

Sketch the graph of \[ y= \dfrac1 { x \ln x} \text{ for \(x>0\), \(x\ne1\)}.\] You may assume that \(x\ln x \to 0\) as \(x\to 0\). The continuous random variable \(X\) has probability density function \[ \f(x) = \begin{cases} \dfrac \lambda {x\ln x}& \text{for \(a\le x \le b\)}\;, \\[3mm] \ \ \ 0 & \text{otherwise }, \end{cases} \] where \(a\), \(b\) and \(\lambda\) are suitably chosen constants.

  1. In the case \(a=1/4\) and \(b=1/2\), find \(\lambda\,\).
  2. In the case \(\lambda=1\) and \(a>1\), show that \(b=a^\e\).
  3. In the case \(\lambda =1\) and \(a=\e\), show that \(\P(\e^{3/2}\le X \le \e^2)\approx \frac {31}{108}\,\).
  4. In the case \(\lambda =1\) and \(a=\e^{1/2}\), find \(\P(\e^{3/2}\le X \le \e^2)\;\).

Show Solution
  1. \begin{align*} 1 &= \int_{1/4}^{1/2} \frac{\lambda}{x\ln x} \, dx \\ &= \lambda\left [ \ln |\ln x| \right ]_{1/4}^{1/2} \\ &= \lambda \l \ln |-\ln 2| - \ln |-2 \ln 2| \r \\ &= \lambda (-\ln 2) \end{align*} So \(\lambda = -\frac{1}{\ln 2} = \frac{1}{\ln \frac12}\)
  2. \begin{align*} 1 &= \int_{a}^{b} \frac{1}{x\ln x} \, dx \\ &= \left [ \ln |\ln x| \right ]_{a}^{b} \\ &= \l \ln \ln b - \ln \ln a \r \\ &= \ln \l \frac{\ln b}{\ln a} \r \\ \end{align*} So \(b = e^{a}\)
  3. If \(\lambda = 1, a = e, b = e^e\), then \begin{align*} \P(\e^{3/2}\le X \le \e^2) &= \int_{e^{3/2}}^{e^2} \frac{1}{x \ln x} \, dx \\ &= \left [ \ln \ln x \right]_{e^{3/2}}^{e^2} \\ &= \ln 2 - \ln \frac{3}{2} \\ &= \ln \frac{4}{3} \\ &= \ln \l 1 + \frac{1}{3} \r \\ &\approx \frac{1}{3} - \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} - \frac{1}{4 \cdot 3^4} \\ &= \frac{31}{108} \end{align*}
  4. Note that \(2 > e^{\frac12} > 1\) so \(a = e^{\frac12}, b = e^{\frac{e}2}\). Since \(3 > e \Rightarrow e^{3/2} > e^{\frac{e}{2}}\) this probability is out of range, therefore \(\P(\e^{3/2}\le X \le \e^2) = 0\)
2005 Paper 1 Q14
D: 1516.0 B: 1513.9

The random variable \(X\) can take the value \(X=-1\), and also any value in the range \(0\le X <\infty\,\). The distribution of \(X\) is given by \[ \P(X=-1) =m \,, \ \ \ \ \ \ \ \P(0\le X\le x) = k(1-\e^{-x})\,, \] for any non-negative number \(x\), where \(k\) and \(m\) are constants, and \(m <\frac12\,\).

  1. Find \(k\) in terms of \(m\).
  2. Show that \(\E(X)= 1-2m\,\).
  3. Find, in terms of \(m\), \(\var (X)\) and the median value of \(X\).
  4. Given that \[ \int_0^\infty y^2 \e^{-y^2} \d y = \tfrac14 \sqrt{ \pi}\;,\] find \(\E\big(\vert X \vert^{\frac12}\big)\,\) in terms of \(m\).

Show Solution
  1. We must have the total probability summing to \(1\), therefore \(1 =m + k\) (as \(x \to \infty\)) therefore \(k = 1-m\).
  2. \(\,\) \begin{align*} && \E[X] &= \mathbb{P}(X=-1) \cdot (-1) + \int_0^{\infty} kx e^{-x} \d x \\ &&&= -m + (1-m) = 1-2m \end{align*}
  3. \(\,\) \begin{align*} && \var[X] &= \E[X^2]-\E[X]^2 \\ &&&= \mathbb{P}(X=-1)\cdot(-1)^2 + (1-m)\int_0^{\infty} x^2e^{-x} \d x - (1-2m)^2 \\ &&&= m + (1-m)(1+1^2) - (1-2m)^2 \\ &&&= 3-4m - 1+4m -4m^2 \\ &&&= 2(1-m^2) \end{align*} To find the median \(q\), we need \begin{align*} && \frac12 &= \mathbb{P}(X \leq q) \\ &&&= m + (1-m)(1-e^{-q}) \\ \Rightarrow && e^{-q} &= 1-\frac{\frac12-m}{1-m} \\ &&&= \frac{1-m - \frac12+m}{1-m} \\ &&&= \frac{1}{2(1-m)} \\ \Rightarrow && q &= \ln 2(1-m) \end{align*}
  4. \(\,\) \begin{align*} && \E\left [|X|^{\frac12}\right] &= \mathbb{P}(X=-1) \cdot 1 + \int_0^{\infty} \sqrt{x} (1-m)e^{-x} \d x \\ &&&= m + (1-m)\int_0^\infty \sqrt{x} e^{-x} \d x \\ u^2 = x, \d x = 2u \d u : &&&= m + (1-m) \int_{u=0}^{u=\infty} u e^{-u^2} \cdot 2u \d u \\ &&&= m + 2(1-m) \int_0^{\infty} u^2 e^{-u^2} \d u \\ &&&= m + (1-m)\frac{\sqrt{\pi}}2 \end{align*}
2001 Paper 2 Q13
D: 1600.0 B: 1517.3

The life times of a large batch of electric light bulbs are independently and identically distributed. The probability that the life time, \(T\) hours, of a given light bulb is greater than \(t\) hours is given by \[ \P(T>t) \; = \; \frac{1}{(1+kt)^\alpha}\;, \] where \(\alpha\) and \(k\) are constants, and \(\alpha >1\). Find the median \(M\) and the mean \(m\) of \(T\) in terms of \(\alpha\) and \(k\). Nine randomly selected bulbs are switched on simultaneously and are left until all have failed. The fifth failure occurs at 1000 hours and the mean life time of all the bulbs is found to be 2400 hours. Show that \(\alpha\approx2\) and find the approximate value of \(k\). Hence estimate the probability that, if a randomly selected bulb is found to last \(M\) hours, it will last a further \(m-M\) hours.

Show Solution
The median \(M\) is the value such that \begin{align*} && \frac12 &= \mathbb{P}(T > M) \\ &&&= \frac1{(1+kM)^\alpha} \\ \Rightarrow && 2 &= (1+kM)^{\alpha} \\ \Rightarrow && M &= \frac{2^{1/\alpha}-1}{k} \end{align*} The distribution of \(T\) is \(f_T(t) = \frac{k \alpha}{(1+kt)^{\alpha+1}}\) and so \begin{align*} && m &= \int_0^\infty t f_T(t) \d t \\ &&&= \int_0^\infty \frac{tk \alpha}{(1+kt)^{\alpha+1}} \d t \\ &&&= \int_0^\infty \frac{\alpha+tk \alpha-\alpha}{(1+kt)^{\alpha+1}} \d t \\ &&&= \alpha \int_0^\infty (1+kt)^{-\alpha} \d t - \alpha \int_0^\infty (1+kt)^{-(\alpha+1)} \d t \\ &&&= \alpha \left [ -\frac1{k(\alpha-1)}(1+kt)^{-\alpha+1}\right]_0^\infty- \alpha \left [ -\frac1{k\alpha}(1+kt)^{-\alpha}\right]_0^\infty \\ &&&= \frac{\alpha}{k(\alpha-1)} - \frac{1}{k} \\ &&&= \frac{1}{k(\alpha-1)} \end{align*} \begin{align*} && \frac{2^{1/\alpha}-1}{k} &= 1000 \\ && \frac{1}{k(\alpha-1)} &= 2400 \\ \Rightarrow && \frac{\alpha-1}{2^{1/\alpha}-1} &\approx 2.4 \\ && \frac{2-1}{\sqrt2-1} &= \sqrt{2}+1 \approx 2.4 \\ \Rightarrow && \alpha &\approx 2 \\ && k &= \frac{1}{2400} \end{align*} \begin{align*} && \mathbb{P}(T > m | T > M) &= \frac{\mathbb{P}(T > m)}{\mathbb{P}(T > M)} \\ &&&= \frac{2}{(1+km)^{\alpha}} \\ &&&= \frac{2}{(1 + \frac{1}{\alpha-1})^\alpha} \\ &&&\approx \frac{2}{4} =\frac12 \end{align*}
1999 Paper 2 Q13
D: 1600.0 B: 1484.0

A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio, \(R\), of the length of the shorter piece to the length of the longer piece is less than \(r\). Find the probability density function for \(R\), and calculate the mean and variance of \(R\).

Show Solution
Let \(X \sim U[0, \tfrac12]\) be the shorter piece, so \(R = \frac{X}{1-X}\), and \begin{align*} && \mathbb{P}(R \leq r) &= \mathbb{P}(\tfrac{X}{1-X} \leq r) \\ &&&= \mathbb{P}(X \leq r - rX) \\ &&&= \mathbb{P}((1+r)X \leq r) \\ &&&= \mathbb{P}(X \leq \tfrac{r}{1+r} ) \\ &&&= \begin{cases} 0 & r < 0 \\ \frac{2r}{1+r} & 0 \leq r \leq 1 \\ 1 & r > 1 \end{cases} \\ \\ && f_R(r) &= \begin{cases} \frac{2}{(1+r)^2} & 0 \leq r \leq 1 \\ 0 & \text{otherwise} \end{cases} \end{align*} Let \(Y \sim U[\tfrac12, 1]\) be the longer piece, then \(R = \frac{1-Y}{Y} = Y^{-1} - 1\) and \begin{align*} \E[R] &= \int_{\frac12}^1 (y^{-1}-1) 2 \d y \\ &= 2\left [\ln y - y \right]_{\frac12}^1 \\ &= -2 + 2\ln2 +2\frac12 \\ &= 2\ln2 -1 \\ \\ \E[R^2] &= \int_{\frac12}^1 (y^{-1}-1)^2 2 \d y\\ &= 2\left [-y^{-1} -2\ln y + 1 \right]_{\frac12}^1 \\ &= 2 \left ( 2 - 2\ln 2+\frac12\right) \\ &= 3-4\ln 2 \\ \var[R] &= 3 - 4 \ln 2 -(2\ln 2-1)^2 \\ &= 2 - 4(\ln 2)^2 \end{align*}
1994 Paper 1 Q14
D: 1500.0 B: 1532.7

Each of my \(n\) students has to hand in an essay to me. Let \(T_{i}\) be the time at which the \(i\)th essay is handed in and suppose that \(T_{1},T_{2},\ldots,T_{n}\) are independent, each with probability density function \(\lambda\mathrm{e}^{-\lambda t}\) (\(t\geqslant0\)). Let \(T\) be the time I receive the first essay to be handed in and let \(U\) be the time I receive the last one.

  1. Find the mean and variance of \(T_{i}.\)
  2. Show that \(\mathrm{P}(U\leqslant u)=(1-\mathrm{e}^{-\lambda u})^{n}\) for \(u\geqslant0,\) and hence find the probability density function of \(U\).
  3. Obtain \(\mathrm{P}(T>t),\) and hence find the probability density function of \(T\).
  4. Write down the mean and variance of \(T\).

Show Solution
  1. \(T_i \sim \textrm{Exp}(\lambda)\) so \(\E[T_i] = \lambda^{-1}, \var[T_i] = \lambda^{-2}\)
  2. \(\,\) \begin{align*} && \mathbb{P}(U \leq u) &= \mathbb{P}(T_i \leq u\quad \forall i) \\ &&&= \prod \mathbb{P}(T_i \leq u) \\ &&&= \prod \int_0^u \lambda e^{-\lambda t} \d t \\ &&&= (1-e^{-\lambda u})^n \\ \\ \Rightarrow && f_U(u) &= n\lambda e^{-\lambda u}(1-e^{-\lambda u})^{n-1} \end{align*}
  3. \(\,\) \begin{align*} && \mathbb{P}(T > t) &= \mathbb{P}(T_i > t \quad \forall i) \\ &&&= \prod \mathbb{P}(T_i > t) \\ &&&= e^{-n\lambda t} \\ \Rightarrow && f_T(t) &= n\lambda e^{-n\lambda t} \end{align*}
  4. Therefore \(\E[T] = \frac{1}{n\lambda}, \var[T] = \frac{1}{(n\lambda)^2}\)
1993 Paper 1 Q14
D: 1500.0 B: 1505.6

When he sets out on a drive Mr Toad selects a speed \(V\) kilometres per minute where \(V\) is a random variable with probability density \[ \alpha v^{-2}\mathrm{e}^{-\alpha v^{-1}} \] and \(\alpha\) is a strictly positive constant. He then drives at constant speed, regardless of other drivers, road conditions and the Highway Code. The traffic lights at the Wild Wood cross-roads change from red to green when Mr Toad is exactly 1 kilometre away in his journey towards them. If the traffic light is green for \(g\) minutes, then red for \(r\) minutes, then green for \(g\) minutes, and so on, show that the probability that he passes them after \(n(g+r)\) minutes but before \(n(g+r)+g\) minutes, where \(n\) is a positive integer, is \[ \mathrm{e}^{-\alpha n(g+r)}-\mathrm{e}^{-\alpha\left(n(g+r)\right)+g}. \] Find the probability \(\mathrm{P}(\alpha)\) that he passes the traffic lights when they are green. Show that \(\mathrm{P}(\alpha)\rightarrow1\) as \(\alpha\rightarrow\infty\) and, by noting that \((\mathrm{e}^{x}-1)/x\rightarrow1\) as \(x\rightarrow0\), or otherwise, show that \[ \mathrm{P}(\alpha)\rightarrow\frac{g}{r+g}\quad\mbox{ as }\alpha\rightarrow0. \] {[}NB: the traffic light show only green and red - not amber.{]}

1989 Paper 1 Q14
D: 1516.0 B: 1453.5

The prevailing winds blow in a constant southerly direction from an enchanted castle. Each year, according to an ancient tradition, a princess releases 96 magic seeds from the castle, which are carried south by the wind before falling to rest. South of the castle lies one league of grassy parkland, then one league of lake, then one league of farmland, and finally the sea. If a seed falls on land it will immediately grow into a fever tree. (Fever trees do not grow in water). Seeds are blown independently of each other. The random variable \(L\) is the distance in leagues south of the castle at which a seed falls to rest (either on land or water). It is known that the probability density function \(\mathrm{f}\) of \(L\) is given by \[ \mathrm{f}(x)=\begin{cases} \frac{1}{2}-\frac{1}{8}x & \mbox{ for }0\leqslant x\leqslant4,\\ 0 & \mbox{ otherwise.} \end{cases} \] What is the mean number of fever trees which begin to grow each year?

  1. The random variable \(Y\) is defined as the distance in leagues south of the castle at which a new fever tree grows from a seed carried by the wind. Sketch the probability density function of \(Y\), and find the mean of \(Y\).
  2. One year messengers bring the king the news that 23 new fever trees have grown in the farmland. The wind never varies, and so the king suspects that the ancient tradition have not been followed properly. Is he justified in his suspicions?

Show 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: $f_Y(t) = \begin{cases} ( \frac45 - \frac15t ) & \text{if } t \in [0,1]\cup[2,3] \\ 0 & \text{otherwise} \end{cases}$
    TikZ diagram
    \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.
1988 Paper 1 Q16
D: 1500.0 B: 1498.6

Wondergoo is applied to all new cars. It protects them completely against rust for three years, but thereafter the probability density of the time of onset of rust is proportional to \(t^{2}/(1+t^{2})^{2}\) for a car of age \(3+t\) years \((t\geqslant0)\). Find the probability that a car becomes rusty before it is \(3+t\) years old. Every car is tested for rust annually on the anniversary of its manufacture. If a car is not rusty, it will certainly pass; if it is rusty, it will pass with probability \(\frac{1}{2}.\) Cars which do not pass are immediately taken off the road and destroyed. What is the probability that a randomly selected new car subsequently fails a test taken on the fifth anniversary of its manufacture? Find also the probability that a car which was destroyed immediately after its fifth anniversary test was rusty when it passed its fourth anniversary test.

Show Solution
Given the probability density after \(3\) years is proportional to \(\frac{t^2}{(1+t^2)^2}\) then we must have that: \begin{align*} && 1 &= A \int_0^{\infty} \frac{t^2}{(1+t^2)^2} \, \d t \\ &&&= A \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{\infty} + \frac{A}2 \int_0^{\infty} \frac{1}{1+t^2} \d t \\ &&&= \frac{A}{2} \frac{\pi}{2} \\ \Rightarrow && A &= \frac{4}{\pi} \end{align*} In order to fail a test on the fifth anniversary, there are two possibilities for when we went faulty. We could have gone faulty before \(4\) years, got lucky once and then failed the second test, or gone faulty in the next year and then failed the first test. \begin{align*} \P(\text{rusty before } 4 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\ &= \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{1} + \frac{2}{\pi} \int_0^{1} \frac{1}{1+t^2} \d t \\ &= -\frac{1}{\pi} + \frac{2}{\pi} \frac{\pi}{4} \\ &= \frac12 - \frac{1}{\pi} \\ &\approx 0.181690\cdots \\ \\ \P(\text{rusty before } 5 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\ &= \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{2} + \frac{2}{\pi} \int_0^{2} \frac{1}{1+t^2} \d t \\ &= -\frac{4}{5\pi} + \frac{2}{\pi} \tan^{-1} 2 \\ &\approx 0.450184\cdots \\ \end{align*} Therefore: \begin{align*} \P(\text{fails 5th anniversary}) &= \P(\text{rusty before } 4 \text{ years}) \P(\text{pass one, fail other}) + \\ & \quad \quad + \P(\text{rusty between 4 and 5 years}) \P(\text{fail}) \\ &= 0.181690\cdots \cdot \frac{1}{4} + \frac{1}{2} ( 0.450184\cdots- 0.181690\cdots) \\ &= \frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots \\ &= 0.1796688\cdots \\ &= 18.0\%\,\, (3\text{ s.f.}) \end{align*} We also must have that: \begin{align*} \P(\text{rusty at 4 years}|\text{destroyed at 5}) &= \frac{\P(\text{rusty at 4 years and destroyed at 5})}{\P(\text{destroyed at 5})} \\ &= \frac{0.181690\cdots \cdot \frac{1}{4}}{\frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots} \\ &= 0.252811\cdots \\ &= 25.3\%\,\,(3\text{ s.f.}) \end{align*}