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2019 Paper 2 Q12
D: 1500.0 B: 1500.0
probability continuous probability distribution probability density function expectation variance interquartile range binomial expansion median inequality power function distribution

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 $$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots < 4.$$

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*}

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2012 Paper 1 Q12
D: 1484.0 B: 1516.0
probability conditional probability Bayes' theorem continuous probability distribution PDF integration fire extinguisher testing law of total probability

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.

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*}

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2011 Paper 1 Q13
D: 1484.0 B: 1471.5
probability continuous probability distribution probability density function integration mean quantile piecewise function arcsin

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 \(a\).

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*}

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1988 Paper 1 Q16
D: 1500.0 B: 1498.6
probability continuous probability distribution probability density function conditional probability Bayes' theorem integration trigonometric substitution rust/decay model

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.

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*}

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