28 problems found
It has been observed that Professor Ecks proves three types of theorems: 1, those that are correct and new; 2, those that are correct, but already known; 3, those that are false. It has also been observed that, if a certain of her theorems is of type \(i\), then her next theorem is of type \(j\) with probability \(p\low_{ij},\) where \(p\low_{ij}\) is the entry in the \(i\)th row and \(j\)th column of the following array: \[ \begin{pmatrix}0.3 & 0.3 & 0.4\\ 0.2 & 0.4 & 0.4\\ 0.1 & 0.3 & 0.6 \end{pmatrix}\,. \] Let \(a_{i},\) \(i=1,2,3\), be the probability that a given theorem is of type \(i\), and let \(b_{j}\) be the consequent probability that the next theorem is of type \(j\).
A message of \(10^{k}\) binary digits is sent along a fibre optic cable with high probabilities \(p_{0}\) and \(p_{1}\) that the digits 0 and 1, respectively, are received correctly. If the probability of a digit in the original message being a 1 is \(\alpha,\) find the probability that the entire message is received correctly. Find the probability \(\beta\) that a randomly chosen digit in the message is received as a 1 and show that \(\beta=\alpha\) if, and only if \[ \alpha=\frac{q_{0}}{q_{1}+q_{0}}, \] where \(q_{0}=1-p_{0}\) and \(q_{1}=1-p_{1}.\) If this condition is satisfied and the received message consists entirely of zeros, what is the probability that it is correct? If now \(q_{0}=q_{1}=q\) and \(\alpha=\frac{1}{2},\) find the approximate value of \(q\) which will ensure that a message of one million binary digits has a fifty-fifty chance of being received entirely correctly. The probability of error \(q\) is proportional to the square of the length of the cable. Initially the length is such that the probability of a message of one million binary bits, among which 0 and 1 are equally likely, being received correctly is \(\frac{1}{2}.\) What would this probability become if a booster station were installed at its mid-point, assuming that the booster station re-transmits the received version of the message, and assuming that terms of order \(q^{2}\) may be ignored?
There are 28 colleges in Cambridge, of which two (New Hall and Newnham) are for women only; the others admit both men and women. Seven women, Anya, Betty, Celia, Doreen, Emily, Fariza and Georgina, are all applying to Cambridge. Each has picked three colleges at random to enter on her application form.
Solution:
Captain Spalding is on a visit to the idyllic island of Gambriced. The population of the island consists of the two lost tribes of Frodox and the latest census shows that \(11/16\) of the population belong to the Ascii who tell the truth \(3/4\) of the time and \(5/16\) to the Biscii who always lie. The answers of an Ascii to each question (even if it is the same as one before) are independent. Show that the probability that an Ascii gives the same answer twice in succession to the same question is \(5/8\). Show that the probability that an Ascii gives the same answer twice is telling the truth is \(9/10.\) Captain Spalding addresses one of the natives as follows. \hspace{1.5em} \textsl{Spalding: }My good man, I'm afraid I'm lost. Should I go left or right to reach the nearest town?\nolinebreak \hspace{1.5em}\textsl{Native: }Left. \hspace{1.5em}\textsl{Spalding: }I am a little deaf. Should I go left or right to reach the nearest town? \hspace{1.5em}\textsl{Native (patiently): }Left. Show that, on the basis of this conversation, Captain Spalding should go left to try and reach the nearest town and that there is a probability \(99/190\) that this is the correct direction. The conversation resumes as follows. \hspace{1.5em}\textsl{Spalding: }I'm sorry I didn't quite hear that. Should I go left or right to reach the nearest town? \hspace{1.5em}\textsl{Native (loudly and clearly): }Left. Shouls Captain Spalding go left or right and why? Show that if he follows your advice the probability that this is the correct direction is \(331/628\).
A taxi driver keeps a packet of toffees and a packet of mints in her taxi. From time to time she takes either a toffee (with probability \(p\)) or mint (with probability \(q=1-p\)). At the beginning of the week she has \(n\) toffees and \(m\) mints in the packets. On the \(N\)th occasion that she reaches for a sweet, she discovers (for the first time) that she has run out of that kind of sweet. What is the probability that she was reaching for a toffee?
Solution: \begin{align*} \mathbb{P}(\text{run out reading for toffee on } N\text{th occassion}) &= \binom{N-1}{n}p^nq^{N-1-n}p \end{align*} Since out of the first \(N-1\) times, we need to choose toffee \(n\) times, and then choose it again for the \(N\)th time. Therefore: \begin{align*} \mathbb{P}(\text{reaching for toffee} | \text{run out on }N\text{th occassion}) &= \frac{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion})}{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion}) + \mathbb{P}(\text{reaching for mint and run out on }N\text{th occassion})} \\ &= \frac{ \binom{N-1}{n}p^nq^{N-1-n}p}{ \binom{N-1}{n}p^nq^{N-1-n}p + \binom{N-1}{m}q^mp^{N-1-m}q} \\ &= \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} \l \frac{q}{p} \r^{m+ n+ 2-N}} \end{align*} Some conclusions we can draw from this are: As \(p \to 1, q \to 0\), the probability they were reaching for a Toffee tends to \(1\). (And vice versa). If \(p = q\), then the probability is: \begin{align*} \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} } \end{align*} Since \(n+1 \leq N \leq n+m+1\) where \(n \geq m\) we can notice that: \begin{align*} \text{if } N = m + n + 1 && \binom{m+n+1 - 1}{n} &= \binom{m+n+1 - 1}{m} & \text{ so } \mathbb{P} = \frac12 \\ \text{if } N = n+k && \binom{n+k-1}{n} &< \binom{n+k-1}{m} & \text{ so } \mathbb{P} < \frac12 \\ \end{align*}
The probability that there are exactly \(n\) misprints in an issue of a newspaper is \(\mathrm{e}^{-\lambda}\lambda^{n}/n!\) where \(\lambda\) is a positive constant. The probability that I spot a particular misprint is \(p\), independent of what happens for other misprints, and \(0 < p < 1.\)
Solution:
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*}
Each day, I choose at random between my brown trousers, my grey trousers and my expensive but fashionable designer jeans. Also in my wardrobe, I have a black silk tie, a rather smart brown and fawn polka-dot tie, my regimental tie, and an elegant powder-blue cravat which I was given for Christmas. With my brown or grey trousers, I choose ties (including the cravat) at random, except of course that I don\textquoteright t wear the cravat with the brown trousers or the polka-dot tie with the grey trousers. With the jeans, the choice depends on whether it is Sunday or one of the six weekdays: on weekdays, half the time I wear a cream-coloured sweat-shirt with \(E=mc{}^{2}\) on the front and no tie; otherwise, and on Sundays (when naturally I always wear a tie), I just pick at random from my four ties. This morning, I received through the post a compromising photograph of myself. I often receive such photographs and they are equally likely to have been taken on any day of the week. However, in this particular photograph, I am wearing my black silk tie. Show that, on the basis of this information, the probability that the photograph was taken on Sunday is \(11/68\). I should have mentioned that on Mondays I lecture on calculus and I therefore always wear my jeans (to make the lectures seem easier to understand). Find, on the basis of the complete information, the probability that the photograph was taken on Sunday. [The phrase `at random' means `with equal probability'.]
I can choose one of three routes to cycle to school. Via Angle Avenue the distance is 5\(\,\)km, and I am held up at a level crossing for \(A\) minutes, where \(A\) is a continuous random variable uniformly distributed between \(0\) and 10. Via Bend Boulevard the distance is 4\(\,\)km, and I am delayed, by talking to each of \(B\) friends for 3\(\,\)minutes, for a total of \(3B\) minutes, where \(B\) is a random variable whose distribution is Poisson with mean 4. Via Detour Drive the distance should be only 2\(\,\)km, but in addition, due to never-ending road works, there are five places at each of which, with probability \(\frac{4}{5},\) I have to make a detour that increases the distance by 1\(\,\)km. Except when delayed by talking to friends or at the level crossing, I cycle at a steady 12\(\,\)km\(\,\)h\(^{-1}\). For each of the three routs, calculate the probability that a journey lasts at least 27 minutes. Each day I choose one of the three routes at random, and I am equally likely to choose any of the three alternatives. One day I arrive at school after a journey of at least 27 minutes. What is the probability that I came via Bend Boulevard? Which route should I use all the time: \begin{questionparts} \item if I wish my average journey time to be as small as possible; \item if I wish my journey time to be less than 32 minutes as often as possible? \end{questionpart} Justify your answers.
Solution: \(A \sim 5\cdot 5 + U[0,10]\) \(B \sim 4 \cdot 5 + 3 \textrm{Po}(4)\) \(C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5\) \begin{align*} && \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\ && \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\ &&&= \mathbb{P}(Po(4) \leq 2) \\ &&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\ &&&= 0.23810\ldots \\ && \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\ &&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\ &&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\ &&&= 0.26272 \end{align*} \begin{align*} \mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\ &= \frac{\frac13 \cdot 0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot 0.23810\ldots + \frac13 \cdot 0.26272} \\ &= 0.3397\ldots \\ &= 0.340 \, \, (3\text{ s.f.}) \end{align*}
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*}
The parliament of Laputa consists of 60 Preservatives and 40 Progressives. Preservatives never change their mind, always voting the same way on any given issue. Progressives vote at random on any given issue.
Solution:
My two friends, who shall remain nameless, but whom I shall refer to as \(P\) and \(Q\), both told me this afternoon that there is a body in my fridge. I'm not sure what to make of this, because \(P\) tells the truth with a probability of only \(p\), while \(Q\) (independently) tells the truth with probability \(q\). I haven't looked in the fridge for some time, so if you had asked me this morning, I would have said that there was just as likely to be a body in it as not. Clearly, in view of what \(P\) and \(Q\) told me, I must revise this estimate. Explain carefully why my new estimate of the probability of there being a body in the fridge should be \[ \frac{pq}{1-p-q+2pq}. \] I have now been to look in the fridge, and there is indeed a body in it; perhaps more than one. It seems to me that only my enemy \(A\), or my enemy \(B\), or (with a bit of luck) both \(A\) and \(B\) could be in my fridge, and this morning I would have judged these three possibilities to be equally likely. But tonight I asked \(P\) and \(Q\) separately whether or not \(A\) was in the fridge, and they each said that he was. What should be my new estimate of the probability that both \(A\) and \(B\) are in my fridge? Of course, I tell the truth always.
Solution: \begin{align*} \mathbb{P}(\text{body in fridge} | \text{P and Q say so}) &= \frac{\mathbb{P}(\text{body in fridge and P and Q say so})}{\mathbb{P}(\text{P and Q say so})} \\ &= \frac{\frac12 pq}{\mathbb{P}(\text{body in fridge and P and Q say so})+\mathbb{P}(\text{no body in fridge and P and Q say so})} \\ &= \frac{\frac12 pq}{\frac12 pq + \frac12(1-p)(1-q)} \\ &= \frac{pq}{pq + 1-p-q+pq} \\ &= \frac{pq}{1-p-q+2pq} \end{align*} \begin{align*} \mathbb{P}(\text{A and B in fridge} | \text{P and Q say A is in fridge}) &= \frac{\mathbb{P}(\text{A and B in fridge and P and Q say A is in fridge}) }{\mathbb{P}( \text{P and Q say A is in fridge}) } \\ &= \frac{\frac13pq}{\frac13pq+\frac13pq+\frac13(1-p)(1-q)} \\ &= \frac{pq}{1-p-q+3pq} \end{align*}
A patient arrives with blue thumbs at the doctor's surgery. With probability \(p\) the patient is suffering from Fenland fever and requires treatment costing \(\pounds 100.\) With probability \(1-p\) he is suffering from Steppe syndrome and will get better anyway. A test exists which infallibly gives positive results if the patient is suffering from Fenland fever but also has probability \(q\) of giving positive results if the patient is not. The test cost \(\pounds 10.\) The doctor decides to proceed as follows. She will give the test repeatedly until either the last test is negative, in which case she dismisses the patient with kind words, or she has given the test \(n\) times with positive results each time, in which case she gives the treatment. In the case \(n=0,\) she treats the patient at once. She wishes to minimise the expected cost \(\pounds E_{n}\) to the National Health Service.
Solution: