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2005 Paper 3 Q13
A pack of cards consists of \(n+1\) cards, which are printed with the integers from \(0\) to \(n\). A~game consists of drawing cards repeatedly at random from the pack until the card printed with 0 is drawn, at which point the game ends. After each draw, the player receives \(\pounds 1\) if the card drawn shows any of the integers from \(1\) to \(w\) inclusive but receives nothing if the card drawn shows any of the integers from \(w+1\) to \(n\) inclusive.
- In one version of the game, each card drawn is replaced immediately and randomly in the pack. Explain clearly why the probability that the player wins a total of exactly \(\pounds 3\) is equal to the probability of the following event occurring: out of the first four cards drawn which show numbers in the range \(0\) to \(w\), the numbers on the first three are non-zero and the number on the fourth is zero. Hence show that the probability that the player wins a total of exactly \(\pounds 3\) is equal to \(\displaystyle \frac{w^3}{(w+1)^4}\). Write down the probability that the player wins a total of exactly \(\pounds r\) and hence find the expected total win.
- In another version of the game, each card drawn is removed from the pack. Show that the expected total win in this version is half of the expected total win in the other version.
2004 Paper 3 Q13
A men's endurance competition has an unlimited number of rounds. In each round, a competitor has, independently, a probability \(p\) of making it through the round; otherwise, he fails the round. Once a competitor fails a round, he drops out of the competition; before he drops out, he takes part in every round. The grand prize is awarded to any competitor who makes it through a round which all the other remaining competitors fail; if all the remaining competitors fail at the same round the grand prize is not awarded. If the competition begins with three competitors, find the probability that:
- all three drop out in the same round;
- two of them drop out in round \(r\) (with \(r \ge 2\)) and the third in an earlier round;
- the grand prize is awarded.
Solution:
- This is the same as the sum of the probability that they all drop out in the \(k\)th round for all values of \(k\), ie \begin{align*} \mathbb{P}(\text{all drop in same round}) &= \sum_{k=0}^\infty \mathbb{P}(\text{all drop out in the }k+1\text{th round}) \\ &= \sum_{k=0}^{\infty}(p^k(1-p))^3 \\ &= (1-p)^3 \sum_{k=0}^{\infty}p^{3k} \\ &= \frac{(1-p)^3}{1-p^3} \\ &= \frac{1+3p(1-p)(p-(1-p))-p^3}{1-p^3} \\ &= \frac{1-p^3-3p(1-p)(1-2p)}{1-p^3} \end{align*}
- There are \(3\) ways to choose the person who drops out earlier, and then they can drop out in round \(0, 1, \cdots r-1\) \begin{align*} \mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) &= 3\sum_{k=0}^{r-2} (p^{r-1}(1-p))^2p^k(1-p) \\ &= 3p^{2r-2}(1-p)^3 \sum_{k=0}^{r-2}p^k \\ &= 3p^{2r-2}(1-p)^3 \frac{1-p^{r-1}}{1-p} \\ &= 3p^{2r-2}(1-p)^2(1-p^{r-1}) \end{align*}
- The probability exactly \(2\) finish after the third \begin{align*} \mathbb{P}(\text{exactly two drop out after third}) &= \sum_{r=2}^{\infty}\mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) \\ &= \sum_{r=2}^{\infty}3p^{2r-2}(1-p)^2(1-p^{r-1}) \\ &= 3(1-p)^2p^{-2}\sum_{r=2}^{\infty}(p^{2r}-p^{3r-1}) \\ &= 3(1-p)^2p^{-2} \left( \frac{p^4}{1-p^2} - \frac{p^5}{1-p^3} \right) \\ &= \frac{3(1-p)^2(p^2(1-p^3)-p^3(1-p^2))}{(1-p^2)(1-p^3)}\\ &= \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)}\\ \end{align*} Therefore the probability the grand prize is not awarded is \begin{align*} P &= 1 - \frac{(1-p)^3}{1-p^3} - \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\ &= \frac{(1-p^3)(1-p^2) - (1-p)^3(1-p^2)-3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\ &= \frac{(1-p^3)(1-p^2) - (1-p)^3(1+2p^2)}{(1-p^2)(1-p^3)} \\ \end{align*}
2003 Paper 1 Q13
If a football match ends in a draw, there may be a "penalty shoot-out". Initially the teams each take 5 shots at goal. If one team scores more times than the other, then that team wins. If the scores are level, the teams take shots alternately until one team scores and the other team does not score, both teams having taken the same number of shots. The team that scores wins. Two teams, Team A and Team B, take part in a penalty shoot-out. Their probabilities of scoring when they take a single shot are \(p_A\) and \(p_B\) respectively. Explain why the probability \(\alpha\) of neither side having won at the end of the initial \(10\)-shot period is given by $$\alpha =\sum_{i=0}^5\binom{5}{i}^2(1-p_A)^i(1-p_B)^i\,p_A^{5-i}p_B^{5-i}.$$ Show that the expected number of shots taken is \(\displaystyle 10+ \frac{2\alpha}\beta\;,\) where \(\beta=p_A+p_B-2p_Ap_B\,.\)
Solution: Note that in the first \(10\)-short period the number of goals scored by each team is \(B(5, \p_i)\). For them to be equal they must both have scored the same number of goals, ie \begin{align*} && \alpha &= \sum_{i=0}^5 \mathbb{P}(\text{both teams score }5-i) \\ &&&= \sum_{i=0}^5 \binom{5}{i} (1-p_A)^ip_A^{5-i} \binom{5}{i} (1-p_B)^i p_B^{5-i} \\ &&&= \sum_{i=0}^5 \binom{5}{i} ^2(1-p_A)^i (1-p_B)^i p_A^{5-i} p_B^{5-i} \\ \end{align*} Suppose we make it to the end of the shoot out with scores tied. The probability that we finish each round is \(p_A(1-p_B) + p_B(1-p_A)\) (the probability \(A\) wins or \(B\) wins). This is \(p_A + p_B - 2p_Ap_B = \beta\)). Therefore the number of additional rounds is geometric with parameter \(\beta\) and the expected number of rounds is \(\frac{1}{\beta}\). Each round has two shots, and there is a probability \(\alpha\) of this occuring, ie \(\frac{2\alpha}{\beta}\). Added to the \(10\) guaranteed shots we get the desired result
2003 Paper 2 Q14
The probability of throwing a 6 with a biased die is \(p\,\). It is known that \(p\) is equal to one or other of the numbers \(A\) and \(B\) where \(0 < A < B < 1 \,\). Accordingly the following statistical test of the hypothesis \(H_0: \,p=B\) against the alternative hypothesis \(H_1: \,p=A\) is performed. The die is thrown repeatedly until a 6 is obtained. Then if \(X\) is the total number of throws, \(H_0\) is accepted if \(X \le M\,\), where \(M\) is a given positive integer; otherwise \(H_1\) is accepted. Let \({\alpha}\) be the probability that \(H_1\) is accepted if \(H_0\) is true, and let \({\beta}\) be the probability that \(H_0\) is accepted if \(H_1\) is true. Show that \({\beta} = 1- {\alpha}^K,\) where \(K\) is independent of \(M\) and is to be determined in terms of \(A\) and \(B\,\). Sketch the graph of \({\beta}\) against \({\alpha}\,\).
Solution: \(X \sim Geo(p)\). \(\alpha = \mathbb{P}(X > M | p = B) = (1-B)^{M}\) \(\beta = \mathbb{P}(X \leq M | p = A) = 1 - \mathbb{P}(X > M | p = A) = 1 - (1-A)^{M}\) \begin{align*} \ln \alpha &= M \ln(1-B) \\ \ln (1-\beta) &= M \ln(1-A) \\ \frac{\ln \alpha}{\ln (1-\beta)} &= \frac{\ln(1-B)}{\ln(1-A)} \\ \ln(1-\beta) &= \ln \alpha \frac{\ln (1-A)}{\ln(1-B)} \\ \beta &= 1- \alpha^{ \frac{\ln (1-A)}{\ln(1-B)} } \end{align*} and \(K = \frac{\ln (1-A)}{\ln(1-B)} \) Since \(0 < A < B < 1\) we must have that \(0 < 1 - B < 1-A < 1\) and \(\ln(1-B) < \ln(1-A) < 0\) so \(0 < K < 1\)
2003 Paper 3 Q12
Brief interruptions to my work occur on average every ten minutes and the number of interruptions in any given time period has a Poisson distribution. Given that an interruption has just occurred, find the probability that I will have less than \(t\) minutes to work before the next interruption. If the random variable \(T\) is the time I have to work before the next interruption, find the probability density function of \(T\,\). I need an uninterrupted half hour to finish an important paper. Show that the expected number of interruptions before my first uninterrupted period of half an hour or more is \(\e^3-1\). Find also the expected length of time between interruptions that are less than half an hour apart. Hence write down the expected wait before my first uninterrupted period of half an hour or more.
2001 Paper 3 Q12
A bag contains \(b\) black balls and \(w\) white balls. Balls are drawn at random from the bag and when a white ball is drawn it is put aside.
- If the black balls drawn are also put aside, find an expression for the expected number of black balls that have been drawn when the last white ball is removed.
- If instead the black balls drawn are put back into the bag, prove that the expected number of times a black ball has been drawn when the first white ball is removed is \(b/w\,\). Hence write down, in the form of a sum, an expression for the expected number of times a black ball has been drawn when the last white ball is removed.
2000 Paper 1 Q12
I have \(k\) different keys on my key ring. When I come home at night I try one key after another until I find the key that fits my front door. What is the probability that I find the correct key in exactly \(n\) attempts in each of the following three cases?
- At each attempt, I choose a key that I have not tried before but otherwise each choice is equally likely.
- At each attempt, I choose a key from all my keys and each of the \(k\) choices is equally likely.
- At the first attempt, I choose from all my keys and each of the \(k\) choices is equally likely. Thereafter, I choose from the keys that I did not try the previous time but otherwise each choice is equally likely.
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 \(p=1/6\).
1999 Paper 3 Q14
In the basic version of Horizons (H1) the player has a maximum of \(n\) turns, where \(n \ge 1\). At each turn, she has a probability \(p\) of success, where \(0 < p < 1\). If her first success is at the \(r\)th turn, where \(1 \le r \le n\), she collects \(r\) pounds and then withdraws from the game. Otherwise, her winnings are nil. Show that in H1, her expected winnings are $$ p^{-1}\left[1+nq^{n+1}-(n+1)q^n\right]\quad\hbox{pounds}, $$ where \(q=1-p\). The rules of H2 are the same as those of H1, except that \(n\) is randomly selected from a Poisson distribution with parameter \(\lambda\). If \(n=0\) her winnings are nil. Otherwise she plays H1 with the selected \(n\). Show that in H2, her expected winnings are $$ {1 \over p}{\left(1-{\e^{-{\lambda}p}}\right)} -{{\lambda}q}{\e^{-{\lambda}p}} \quad\hbox{pounds}. $$
Solution: \begin{align*} && \E[H1] &= \sum_{r=1}^n r \cdot \mathbb{P}(\text{first success on }r\text{th turn}) \\ &&&= \sum_{r=1}^n r \cdot q^{r-1}p \\ &&&= p\sum_{r=1}^n r q^{r-1} \\ \\ && \frac{1-x^{n+1}}{1-x} &= \sum_{r=0}^n x^r \\ \Rightarrow && \sum_{r=1}^n r x^{r-1} &= \frac{-(n+1)x^n(1-x) +(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} \\ \\ && \E[H1] &= p\sum_{r=1}^n r q^{r-1} \\ &&&= p\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} \\ &&&= p^{-1}(1-(n+1)q^{n} + nq^{n+1}) \end{align*} Not that if \(n =0\) , the formula for \(\E[H1] = 0\). So \begin{align*} && \E[H2] &= \E[\E[H1|n=N]] \\ &&&= p^{-1}\E \left [ 1-(N+1)q^{N} + Nq^{N+1}\right] \\ &&&= p^{-1}\E \left [ 1-((1-q)N+1)q^{N} \right] \\ &&&= p^{-1}\left (1 - p\E[Nq^N] - G_{Po(\lambda)}(q) \right) \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - \E[Nq^N] \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - q\lambda e^{-\lambda(1-q)} \\ &&&= p^{-1}(1-e^{-\lambda p}) - q\lambda e^{-\lambda p} \end{align*}
1997 Paper 2 Q12
The game of Cambridge Whispers starts with the first participant Albert flipping an un-biased coin and whispering to his neighbour Bertha whether it fell `heads' or `tails'. Bertha then whispers this information to her neighbour, and so on. The game ends when the final player Zebedee whispers to Albert and the game is won, by all players, if what Albert hears is correct. The acoustics are such that the listeners have, independently at each stage, only a probability of 2/3 of hearing correctly what is said. Find the probability that the game is won when there are just three players. By considering the binomial expansion of \((a+b)^n+(a-b)^n\), or otherwise, find a concise expression for the probability \(P\) that the game is won when is it played by \(n\) players each having a probability \(p\) of hearing correctly. % Show in particular that, if \(n\) is even, %\(P(n,1/10) = P(n,9/10)\).% How do you explain this apparent anomaly? To avoid the trauma of a lost game, the rules are now modified to require Albert to whisper to Bertha what he hears from Zebedee, and so keep the game going, if what he hears from Zebedee is not correct. Find the expected total number of times that Albert whispers to Bertha before the modified game ends. \noindent [You may use without proof the fact that \(\sum_1^\infty kx^{k-1}=(1-x)^{-2}\) for \(\vert x\vert<1\).]
1997 Paper 3 Q12
- I toss a biased coin which has a probability \(p\) of landing heads and a probability \(q=1-p\) of landing tails. Let \(K\) be the number of tosses required to obtain the first head and let \[ \mathrm{G}(s)=\sum_{k=1}^{\infty}\mathrm{P}(K=k)s^{k}. \] Show that \[ \mathrm{G}(s)=\frac{ps}{1-qs} \] and hence find the expectation and variance of \(K\).
- I sample cards at random with replacement from a normal pack of \(52\). Let \(N\) be the total number of draws I make in order to sample every card at least once. By expressing \(N\) as a sum \(N=N_{1}+N_{2}+\cdots+N_{52}\) of random variables, or otherwise, find the expectation of \(N\). Estimate the numerical value of this expectation, using the approximations \(\mathrm{e}\approx2.7\) and \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\approx0.5+\ln n\) if \(n\) is large.
Solution:
- Let \(N_i\) be the number of draws between the \((i-1)\)th new card and the \(i\)th new card. (Where \(N_1 = 1\)0 then \(N_i \sim K\) with \(p = \frac{53-i}{52}\)). Therefore \begin{align*} \E[N] &= \E[N_1 + \cdots + N_{52}] \\ &= \E[N_1] + \cdots + \E[N_i] + \cdots + \E[N_{52}] \\ &= 1 + \frac{52}{51} + \cdots + \frac{52}{53-k} + \cdots + \frac{52}{1} \\ &= 52 \left (1 + \frac{1}{2} + \cdots + \frac{1}{52} \right) \\ &= 52 \cdot \left ( 1 + \ln 52 \right) \end{align*} Notice that \(2.7 \times 2.7 = 7.29\) and \(7.3 \times 7.3 \approx 53.3\) so \(\ln 52 \approx 4\) and so our number is \(\approx 52 \cdot 4.5 =234\). [The correct answer actual number is 235.9782]
1994 Paper 2 Q13
The makers of Cruncho (`The Cereal Which Cares') are giving away a series of cards depicting \(n\) great mathematicians. Each packet of Cruncho contains one picture chosen at random. Show that when I have collected \(r\) different cards the expected number of packets I must open to find a new card is \(n/(n-r)\) where \(0\leqslant r\leqslant n-1.\) Show by means of a diagram, or otherwise, that \[ \frac{1}{r+1}\leqslant\int_{r}^{r+1}\frac{1}{x}\,\mathrm{d}x\leqslant\frac{1}{r} \] and deduce that \[ \sum_{r=2}^{n}\frac{1}{r}\leqslant\ln n\leqslant\sum_{r=1}^{n-1}\frac{1}{r} \] for all \(n\geqslant2.\) My children will give me no peace until we have the complete set of cards, but I am the only person in our household prepared to eat Cruncho and my spouse will only buy the stuff if I eat it. If \(n\) is large, roughly how many packets must I expect to consume before we have the set?
1994 Paper 3 Q13
During his performance a trapeze artist is supported by two identical ropes, either of which can bear his weight. Each rope is such that the time, in hours of performance, before it fails is exponentially distributed, independently of the other, with probability density function \(\lambda\exp(-\lambda t)\) for \(t\geqslant0\) (and 0 for \(t < 0\)), for some \(\lambda > 0.\) A particular rope has already been in use for \(t_{0}\) hours of performance. Find the distribution for the length of time the artist can continue to use it before it fails. Interpret and comment upon your result. Before going on tour the artist insists that the management purchase two new ropes of the above type. Show that the probability density function of the time until both ropes fail is \[ \mathrm{f}(t)=\begin{cases} 2\lambda\mathrm{e}^{-\lambda t}(1-\mathrm{e}^{-\lambda t}) & \text{ if }t\geqslant0,\\ 0 & \text{ otherwise.} \end{cases} \] If each performance lasts for \(h\) hours, find the probability that both ropes fail during the \(n\)th performance. Show that the probability that both ropes fail during the same performance is \(\tanh(\lambda h/2)\).
Solution: This is the memoryless property of the exponential distribution so it has the same distribution as when \(t = 0\). Let \(T\) be the time the rope fails, then \begin{align*} && \mathbb{P}(T > t | T > t_0) &= \frac{\mathbb{P}(T > t)}{\mathbb{P}(T > t_0)} \\ &&&= \frac{e^{-\lambda t}}{e^{-\lambda t_0}} \\ &&&= e^{-\lambda(t-t_0)} \end{align*} This means that each rope (as long as it hasn't broken) can be considered "as good as new". Suppose \(T_1, T_2 \sim Exp(\lambda)\) are the time to failures for each rope, then \begin{align*} && \mathbb{P}(\max(T_1, T_2) < t) &= \mathbb{P}(T_1 < t, T_2 < t) \\ &&&= (1-e^{-\lambda t})^2 \\ \Rightarrow && f(t) &= 2(1-e^{-\lambda t}) \cdot (\lambda e^{-\lambda t}) \\ &&&= 2\lambda e^{-\lambda t}(1-e^{-\lambda t}) \end{align*} Therefore \(\max(T_1, T_2) \sim Exp(2\lambda)\) and the pdf is as described. \begin{align*} && \mathbb{P}(\text{both fail during the }n\text{th}) &= \left ( \int_{(n-1)h}^{nh} \lambda e^{-\lambda t} \d t \right)^2 \\ &&&=\left (\left [ -e^{-\lambda t}\right]_{(n-1)h}^{nh} \right)^2 \\ &&&= \left ( e^{-\lambda (n-1)h}( 1-e^{-\lambda h}) \right)^2 \\ &&&= e^{-2(n-1)h\lambda}(e^{-\lambda h}-1)^2 \\ \\ && \mathbb{P}(\text{both fail in same performance}) &= \sum_{n=1}^{\infty} \mathbb{P}(\text{both fail during the }n\text{th}) \\ &&&= \sum_{n=1}^{\infty}e^{-2(n-1)h\lambda}(e^{-\lambda h}-1)^2 \\ &&&= (e^{-\lambda h}-1)^2 \frac{1}{1-e^{-2h\lambda}} \\ &&&= \frac{e^{-\lambda h}-1}{1+e^{-h\lambda}} \\ &&&= \tanh(\lambda h/2) \end{align*}
1993 Paper 2 Q16
At the terminus of a bus route, passengers arrive at an average rate of 4 per minute according to a Poisson process. Each minute, on the minute, one bus arrives with probability \(\frac{1}{4},\) independently of the arrival of passengers or previous buses. Just after eight o'clock there is no-one at the bus stop.
- What is the probability that the first bus arrives at \(n\) minutes past 8?
- If the first bus arrives at 8:05, what is the probability that there are \(m\) people waiting for it?
- Each bus can take 25 people and, since it is the terminus, the bus arrive empty. Explain carefully how you would calculate, to two significant figures, the probability that when the first bus arrives it is unable to pick up all the passengers. Your method should need the use of a calculator and standard tables only. There is no need to carry out the calculation.
1991 Paper 1 Q16
At any instant the probability that it is safe to cross a busy road is \(0.1\). A toad is waiting to cross this road. Every minute she looks at the road. If it is safe, she will cross; if it is not safe, she will wait for a minute before attempting to cross again. Find the probability that she eventually crosses the road without mishap. Later on, a frog is also trying to cross the same road. He also inspects the traffic at one minute intervals and crosses if it is safe. Being more impatient than the toad, he may also attempt to cross when it is not safe. The probability that he will attempt to cross when it is not safe is \(n/3\) if \(n\leqslant3,\) where \(n\) minutes have elapsed since he firrst inspected the road. If he attempts to cross when it is not safe, he is run over with probability \(0.8,\) but otherwise he reaches the other side safely. Find the probability that he eventually crosses the road without mishap. What is the probability that both reptiles safely cross the road with the frog taking less time than the toad? If the frog has not arrived at the other side 2 minutes after he began his attempt to cross, what is the probability that the frog is run over (at some stage) in his attempt to cross? \textit{[Once moving, the reptiles spend a negligible time on their attempt to cross the road.]}
Solution: Since the toad never crosses when it's not safe, she is certain to cross. (Probability she hasn't crossed after the \(n\)th minute is \(0.9^n \to 0\)). \begin{array}{c|c|c|c|c|c|c|c} \text{will try dangerously} & \text{is safe} & \text{has tried} & \text{tries safely} & \text{tries unsafely} & \text{succeeds} & \text{succeeds unsafely} & \text{fails} \\ \hline 0 & 0.1 & 0 & 0.1 & 0 & 0.1 & 0 & 0\\ \frac13 & 0.1 & 0.1 & 0.09 & 0.27 & 0.144 & 0.054 & 0.216\\ \frac23 & 0.1 & 0.46 & 0.054 & 0.324 & 0.1188 & 0.0648 & 0.2592\\ 1 & 0.1 & 0.838 & 0.0162 & 0.1458 & 0.04536 & 0.02916 & 0.11664\\ \hline & & & & & 0.40816 & 0.14796 & \\ \hline \end{array} So \(\mathbb{P}(\text{frog crosses safely}) = 0.40816\) and \(\mathbb{P}(\text{frog beats toad across}) = 0.14796\). \begin{align*} \mathbb{P}(\text{frog run over} | \text{frog not crossed after 2 minutes}) &= \frac{\mathbb{P}(\text{frog run over and frog not crossed after 2 minutes})}{\mathbb{P}(\text{frog not crossed after 2 minutes})} \\ &= \frac{\mathbb{P}(\text{frog run over within 2 minutes})}{\mathbb{P}(\text{frog not crossed after 2 minutes})} \\ &= \frac{\mathbb{P}(\text{frog run over within 2 minutes})}{1-\mathbb{P}(\text{crossed after 2 minutes})} \\ &= \frac{0.216+0.2592}{1-0.3628} \\ &= 0.7457\ldots \end{align*}