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2000 Paper 3 Q12
In a lottery, any one of \(N\) numbers, where \(N\) is large, is chosen at random and independently for each player by machine. Each week there are \(2N\) players and one winning number is drawn. Write down an exact expression for the probability that there are three or fewer winners in a week, given that you hold a winning ticket that week. Using the fact that $$ {\biggl( 1 - {a \over n} \biggr) ^n \approx \e^{-a}}$$ for \(n\) much larger than \(a\), or otherwise, show that this probability is approximately \({2 \over 3}\) . Discuss briefly whether this probability would increase or decrease if the numbers were chosen by the players. Show that the expected number of winners in a week, given that you hold a winning ticket that week, is \( 3-N^{-1}\).
2000 Paper 3 Q13
A set of \(n\) dice is rolled repeatedly. For each die the probability of showing a six is \(p\). Show that the probability that the first of the dice to show a six does so on the \(r\)th roll is $$q^{n r } ( q^{-n} - 1 )$$ where \(q = 1 - p\). Determine, and simplify, an expression for the probability generating function for this distribution, in terms of \(q\) and \(n\). The first of the dice to show a six does so on the \(R\)th roll. Find the expected value of \(R\) and show that, in the case \(n = 2\), \(p=1/6\), this value is \(36/11\). Show that the probability that the last of the dice to show a six does so on the \(r\)th roll is \[ \big(1-q^r\big)^n-\big(1-q^{r-1}\big)^n. \] Find, for the case \(n = 2\), the probability generating function. The last of the dice to show a six does so on the \(S\)th roll. Find the expected value of \(S\) and evaluate this when \(p=1/6\).
2000 Paper 3 Q14
The random variable \(X\) takes only the values \(x_1\) and \(x_2\) (where \( x_1 \not= x_2 \)), and the random variable \(Y\) takes only the values \(y_1\) and \(y_2\) (where \(y_1 \not= y_2\)). Their joint distribution is given by $$ \P ( X = x_1 , Y = y_1 ) = a \ ; \ \ \P ( X = x_1 , Y = y_2 ) = q - a \ ; \ \ \P ( X = x_2 , Y = y_1 ) = p - a \ . $$ Show that if \(\E(X Y) = \E(X)\E(Y)\) then $$ (a - p q ) ( x_1 - x_2 ) ( y_1 - y_2 ) = 0 . $$ Hence show that two random variables each taking only two distinct values are independent if \(\E(X Y) = \E(X) \E(Y)\). Give a joint distribution for two random variables \(A\) and \(B\), each taking the three values \(- 1\), \(0\) and \(1\) with probability \({1 \over 3}\), which have \(\E(A B) = \E( A)\E (B)\), but which are not independent.
Solution: \begin{align*} \mathbb{P}(X = x_1) &= a + q - a = q \\ \mathbb{P}(X = x_2) &= 1 - q \\ \mathbb{P}(Y = y_1) & = a + p - a = p \\ \mathbb{P}(Y = y_2) & = 1 - p \end{align*} \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &= \l qx_1 + (1-q)x_2 \r \l p y_1 + (1-p)y_2\r \\ &= qpx_1y_1 + q(1-p)x_1y_2 + (1-q)px_2y_1 + (1-q)(1-p)x_2y_2 \\ \mathbb{E}(XY) &= ax_1y_1 + (q-a)x_1y_2 + (p-a)x_2y_1 + (1 + a - p - q)x_2y_2 &= \end{align*} Therefore \(\mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)\) is a degree 2 polynomial in the \(x_i, y_i\). If \(x_1 = x_2\) then we have: \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &=x_1 \l p y_1 + (1-p)y_2\r \\ \mathbb{E}(XY) &= x_1(ay_1 + (q-a)y_2 + (p-a)y_1 + (1 + a - p - q)y_2) \\ &= x_1 (py_1 + (1-p)y_2) \end{align*} Therefore \(x_1 - x_2\) is a root and by symmetry \(y_1 - y_2\) is a root. Therefore it remains to check the coefficient of \(x_1y_1\) which is \(a - pq\) to complete the factorisation. For any two random variables taking two distinct values, we can find \(a, q, p\) satisfying the relations above. We also note that \(X\) and \(Y\) are independent if \(\mathbb{P}(X = x_i, Y = y_i) = \mathbb{P}(X = x_i)\mathbb{P}(Y = y_i)\). Since \(x_1 \neq x_2\) and \(y_1 \neq y_2\) and \(\E(A B) = \E( A)\E (B) \Rightarrow a = pq\). But if \(a = pq\), we have \(\mathbb{P}(X = x_1, Y = y_1) = \mathbb{P}(X = x_1)\mathbb{P}(Y = y_1)\) and all the other relations drop out similarly. Consider \begin{align*} \mathbb{P}(A = -1, B = 1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = 0, B = 0) &= \frac{1}{3} \\ \mathbb{P}(A = 1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \end{align*}
1999 Paper 1 Q1
How many integers greater than or equal to zero and less than a million are not divisible by 2 or 5? What is the average value of these integers? How many integers greater than or equal to zero and less than 4179 are not divisible by 3 or 7? What is the average value of these integers?
Solution: There are \(1\,000\,000\) numbers between 1 and a million (inclusive). \(500\,000\) are divisible by \(2\), \(200\,000\) are divisible by \(5\) and \(100\,000\) are divisible by both. Therefore there are: \(1\,000\,000 - 500\,000-200\,000+100\,000 = 400\,000\). (Alternatively, the only numbers are those which are \(1,3,7,9 \pmod{10}\) so there are \(4\) every \(10\), or \(4 \cdot 100\,000\)). We can sum all these values similarly, \begin{align*} S &= \underbrace{\sum_{i=1}^{10^6} i}_{\text{all numbers}}-\underbrace{\sum_{i=1}^{5 \cdot 10^5} 2i}_{\text{all multiples of } 2}-\underbrace{\sum_{i=1}^{2 \cdot 10^5} 5i}_{\text{all multiples of } 5}+\underbrace{\sum_{i=1}^{10^5} 10i}_{\text{all multiples of } 5} \\ &= \frac{10^6 \cdot (10^6 + 1)}{2} - \frac{10^6 \cdot (5\cdot 10^5+1)}{2} - \frac{10^6 \cdot (2\cdot 10^5+1)}{2} + \frac{10^6 \cdot (10^5+1)}{2} \\ &= \frac{10^6 (10^5 \cdot (10-5-2+1))))}{2} \\ &= \frac{10^6 \cdot 10^5 \cdot 4}{2} \\ &= 2\cdot 10^{11} \end{align*} So the average value is \(\frac{2 \cdot 10^{11}}{4 \cdot 10^5} = \frac{10^6}{2} = 500\,000\). (Alternatively, each value can be paired off eg \(999\,999\) with \(1\) and so on, leaving averages of \(500\,000\)). Note that \(4197\) is divisible by \(3\) and \(7\). Using the same long we have: \(4179 - \frac{4179}{3} - \frac{4179}{7} + \frac{4179}{21} = 4179 - 1393 - 597 + 199 = 2388\). The sum will be: \begin{align*} S &= \underbrace{\sum_{i=1}^{4179}i }_{\text{all numbers}}- \underbrace{\sum_{i=1}^{1393}3i }_{\text{multiples of }3}- \underbrace{\sum_{i=1}^{597}7i }_{\text{multiples of }7}+ \underbrace{\sum_{i=1}^{199}21i }_{\text{mulitples of }21} \\ &= \frac{4179 \cdot 4180}{2} - \frac{4179 \cdot 1394}{2} - \frac{4179 \cdot 598}{2} +\frac{4179 \cdot 200}{2} \\ &= \frac{4179 \cdot 2388}{2} \end{align*} So the average value is \(\frac{4179}{2}\).
1999 Paper 1 Q2
A point moves in the \(x\)-\(y\) plane so that the sum of the squares of its distances from the three fixed points \((x_{1},y_{1})\), \((x_{2},y_{2})\), and \((x_{3},y_{3})\) is always \(a^{2}\). Find the equation of the locus of the point and interpret it geometrically. Explain why \(a^2\) cannot be less than the sum of the squares of the distances of the three points from their centroid. [The centroid has coordinates \((\bar x, \bar y)\) where \(3\bar x = x_1+x_2+x_3,\) $3\bar y =y_1+y_2+y_3. $]
Solution: \begin{align*} && a^2 &= d_1^2 + d_2^2 + d_3^2 \\ &&&= (x-x_1)^2+(y-y_1)^2 + (x-x_2)^2+(y-y_2)^2 + (x-x_3)^2+(y-y_3)^2 \\ &&&= \sum (x-\bar{x}+\bar{x}-x_i)^2 + \sum (y-\bar{y}+\bar{y}-y_i)^2 \\ &&&= \sum \left ( (x-\bar{x})^2+(\bar{x}-x_i)^2 + 2(x-\bar{x})(\bar{x}-x_i) \right)+ \sum \left ( (y-\bar{y})^2+(\bar{y}-y_i)^2 + 2(y-\bar{y})(\bar{y}-y_i) \right)\\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2 + 6x\bar{x} -6\bar{x}^2-2x\sum x_i+2\bar{x}\sum x_i + \\ &&&\quad\quad\quad 3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 + 6y\bar{y} -6\bar{y}^2-2y\sum y_i+2\bar{y}\sum y_i \\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2+3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 \\ \\ \Rightarrow && (x-\bar{x})^2+(y-\bar{y})^2 &= \frac13\left ( a^2- \sum \left((\bar{x}-x_i)^2+(\bar{y}-y_i)^2 \right) \right) \end{align*} Therefore the locus is a circle, centre \((\bar{x}, \bar{y})\). radius \(\sqrt{\frac13(a^2 - \text{sum of squares distances of centroid to vertices}})\). \(a^2\) cannot be less than this distance, because clearly the right hand side is always bigger than it!
1999 Paper 1 Q3
The \(n\) positive numbers \(x_{1},x_{2},\dots,x_{n}\), where \(n\ge3\), satisfy $$ x_{1}=1+\frac{1}{x_{2}}\, ,\ \ \ x_{2}=1+\frac{1}{x_{3}}\, , \ \ \ \dots\; , \ \ \ x_{n-1}=1+\frac{1}{x_{n}}\, , $$ and also $$ \ x_{n}=1+\frac{1}{x_{1}}\, . $$ Show that
- \(x_{1},x_{2},\dots,x_{n}>1\),
- \({\displaystyle x_{1}-x_{2}=-\frac{x_{2}-x_{3}}{x_{2}x_{3}}}\),
- \(x_{1}=x_{2}=\cdots=x_{n}\).
1999 Paper 1 Q4
Sketch the following subsets of the \(x\)-\(y\) plane:
- \(|x|+|y|\le 1\) ;
- \(|x-1|+|y-1|\le 1 \) ;
- \(|x-1|-|y+1|\le 1 \) ;
- \(|x|\, |y-2|\le 1\) .
Solution:
1999 Paper 1 Q5
For this question, you may use the following approximations, valid if \(\theta \) is small: \ \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1-\theta^2/2\,\). A satellite \(X\) is directly above the point \(Y\) on the Earth's surface and can just be seen (on the horizon) from another point \(Z\) on the Earth's surface. The radius of the Earth is \(R\) and the height of the satellite above the Earth is \(h\).
- Find the distance \(d\) of \(Z\) from \(Y\) along the Earth's surface.
- If the satellite is in low orbit (so that \(h\) is small compared with \(R\)), show that $$d \approx k(Rh)^{1/2},$$ where \(k\) is to be found.
- If the satellite is very distant from the Earth (so that \(R\) is small compared with \(h\)), show that $$d\approx aR+b(R^2/h),$$ where \(a\) and \(b\) are to be found.
1999 Paper 1 Q6
- Find the greatest and least values of \(bx+a\) for \(-10\leqslant x \leqslant 10\), distinguishing carefully between the cases \(b>0\), \(b=0\) and \(b<0\).
- Find the greatest and least values of \(cx^{2}+bx+a\), where \(c\ge0\), for \(-10\leqslant x \leqslant 10\), distinguishing carefully between the cases that can arise for different values of \(b\) and \(c\).
Solution:
- Case \(b > 0\). Then \(bx+a\) is increasing and the greatest value is \(10b+a\), and the least value \(a-10b\) Case \(b=0\), then \(a\) is constant and the greatest and least value is \(a\) Case \(b < 0\), then \(bx+a\) is decreasing and the greatest value is \(-10b+a\) and the least value is \(10b+a\)
- If \(c = 0\) we have the same cases as above. If \( c > 0\) the consider \(2cx+b\). if \(b-20c > 0\) then our function is increasing on our interval and the greatest value is \(100c+10b+a\) and the least value is \(100c-10b+a\) If \(20c+b < 0\) then our function is decreasing and that calculation is reversed. If neither of these are true, then the minimum will be when \(x = - \frac{b}{2c}\) and the max at one end point.
1999 Paper 1 Q7
Show that \(\sin(k\sin^{-1} x)\), where \(k\) is a constant, satisfies the differential equation $$ (1-x^{2})\frac {\d^2 y}{\d x^2} -x\frac{\d y}{\d x} +k^{2}y=0. \eqno(*) $$ In the particular case when \(k=3\), find the solution of equation \((*)\) of the form \[ y=Ax^{3}+Bx^{2}+Cx+D, \] that satisfies \(y=0\) and \(\displaystyle \frac{\d y}{\d x}=3\) at \(x=0\). Use this result to express \(\sin 3\theta\) in terms of powers of \(\sin\theta\).