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2024 Paper 3 Q12
D: 1500.0 B: 1500.0
cumulative distribution function geometric probability continuous probability distributions integration expected value trigonometric substitution hyperbolic substitution

  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 \(\mathrm{E}(R) = \frac{1}{3}\Bigl(\sqrt{2} + \ln\bigl(\sqrt{2}+1\bigr)\Bigr)\).

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2023 Paper 2 Q12
D: 1500.0 B: 1500.0
continuous probability distributions order statistics probability density function cumulative distribution function median integration maximum of random variables

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 \(\mu(2) < m(2)\).

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2022 Paper 2 Q12
D: 1500.0 B: 1500.0
continuous probability distributions probability density function expectation median mode binomial theorem inequality

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.

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

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2022 Paper 3 Q12
D: 1500.0 B: 1500.0
continuous probability distributions expected value probability density function geometric probability trigonometric identities integration uniform distribution chord length

  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 \(\mathrm{L}(T)\).

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2021 Paper 3 Q11
D: 1500.0 B: 1500.0
exponential distribution continuous probability distributions floor function independence probability density function expectation cumulative distribution function

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.

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.

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2020 Paper 3 Q11
D: 1500.0 B: 1500.0
continuous probability distributions uniform distribution inverse function probability density function expectation variance inequality integration

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 \[ \ln\left(\frac{b-c}{a-c}\right) < \frac{b-a}{c}. \]

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2014 Paper 1 Q13
D: 1500.0 B: 1483.3
continuous probability distributions triangular distribution probability density function piecewise linear function mean median integration cases

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.

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 \(c = \frac{b+a}{2}\)

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2006 Paper 2 Q14
D: 1600.0 B: 1500.0
continuous probability distributions probability density function integration curve sketching logarithmic function substitution normalisation asymptotes

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 \(\P(\e^{3/2}\le X \le \e^2)\;\).

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 \(\P(\e^{3/2}\le X \le \e^2) = 0\)

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2005 Paper 1 Q14
D: 1516.0 B: 1513.9
probability continuous probability distributions mixed distribution expectation variance median exponential distribution substitution absolute value Gaussian integral

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 \(m\).

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*}

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2001 Paper 2 Q13
D: 1600.0 B: 1517.3
continuous probability distributions median mean survival function order statistics parameter estimation conditional probability Pareto distribution memoryless property

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.

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*}

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1994 Paper 1 Q14
D: 1500.0 B: 1532.7
probability continuous probability distributions exponential distribution order statistics CDF PDF mean and variance independence

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 \(T\).

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 \(\E[T] = \frac{1}{n\lambda}, \var[T] = \frac{1}{(n\lambda)^2}\)

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1993 Paper 1 Q14
D: 1500.0 B: 1505.6
probability continuous probability distributions probability density function geometric series exponential distribution limits traffic light problem inverse transform

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.{]}

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1989 Paper 1 Q14
D: 1516.0 B: 1453.5
probability continuous probability distributions probability density function conditional distribution hypothesis testing expected value binomial distribution significance test

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?

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.

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