Each week, a boy receives pocket money only on condition that he wins two games in a row when playing three successive chess games with his father and mother alternating as opponents. The boy knows that his mother's probability of winning is \(\frac{3}{4}\), but his father's probability of winning is only \(\frac{1}{2}\). To maximise his chance of winning two games in succession, should he play the sequence father-mother-father, or mother-father-mother? Assuming that each week the boy plays the sequence more favourable to him, what is the expected number of weeks between two successive occasions on which he receives pocket money?
There are \(k\) distinguishable pairs of shoes in a dark cupboard. A man draws shoes out, one by one, without replacing them. Assume that each possible order of drawing shoes is equally likely.
The number of accidents occurring in a particular year on the M1 motorway has the Poisson distribution with mean \(\lambda_1\), while the number occurring on the M2 has the Poisson distribution with mean \(\lambda_2\). Assuming that the numbers of accidents occurring on different motorways are independent, prove that the total number of accidents on both motorways has the Poisson distribution with mean \(\lambda_1+\lambda_2\). Given that the total number of accidents on the two motorways is \(n\), find the probability that there were \(k\) accidents on the M1.
Solution: Suppose \(X_1 \sim Pois(\lambda_1), X_2 \sim Pois(\lambda_2)\) \begin{align*} \mathbb{P}(X_1+X_2 = n) &= \sum_{i=0}^n \mathbb{P}(X_1 = i, X_2 = n-i) \\ &= \sum_{i=0}^n \mathbb{P}(X_1 = i)\mathbb{P}(X_2 = n-i) \tag{assuming independent} \\ &= \sum_{i=0}^n e^{-\lambda_1} \frac{\lambda_1^i}{i!} e^{-\lambda_2}\frac{\lambda_2^{n-i}}{(n-i)!} \\ &= e^{-(\lambda_1+\lambda_2)} \sum_{i=0}^n \frac{\lambda_1^i\lambda_2^{n-i}}{i!(n-i)!} \\ &= e^{-(\lambda_1+\lambda_2)} \frac{1}{n!}\sum_{i=0}^n \frac{n!\lambda_1^i\lambda_2^{n-i}}{i!(n-i)!} \\ &= e^{-(\lambda_1+\lambda_2)} \frac{1}{n!}\sum_{i=0}^n \binom{n}{i}\lambda_1^i\lambda_2^{n-i} \\ &= e^{-(\lambda_1+\lambda_2)} \frac{1}{n!}(\lambda_1+\lambda_2)^n \end{align*} Therefore their sum has the same distribution as \(Pois(\lambda_1+\lambda_2)\). \begin{align*} \mathbb{P}(X_1 = k | X_1 + X_2 = n) &= \frac{\mathbb{P}(X_1 = k, X_1+X_2 = n)}{\mathbb{P}(X_1+X_2=n)} \\ &= \frac{e^{\lambda_1+\lambda_2}n! }{(\lambda_1+\lambda_2)^n} \mathbb{P}(X_1 = k, X_2 = n-k) \\ &= \frac{e^{\lambda_1+\lambda_2}n! }{(\lambda_1+\lambda_2)^n} \mathbb{P}(X_1 = k)\mathbb{P}(X_2 = n-k) \\ &= \frac{e^{\lambda_1+\lambda_2}n! }{(\lambda_1+\lambda_2)^n}e^{-\lambda_1} \frac{\lambda_1^k}{k!}e^{-\lambda_2}\frac{\lambda_2^{n-k}}{(n-k)!} \\ &= \binom{n}{k} \frac{\lambda_1^k\lambda_2^{n-k}}{(\lambda_1+\lambda_2)^n} \\ &= \binom{n}{k}p^k(1-p)^{n-k} \end{align*} Where \(p = \frac{\lambda_1}{\lambda_1+\lambda_2}\), ie it is distributed \(Binomial(n, \frac{\lambda_1}{\lambda_1+\lambda_2})\)
A tug-of-war contest is to be held between two colleges. The weights of students in College \(A\) follow a normal distribution with mean 140 lb and standard deviation 8 lb. Thanks to the superiority of its kitchens, the weights of students in College \(B\) follow a normal distribution with mean 150 lb and standard deviation 6 lb. Teams are chosen by selecting \(n\) students at random from each college. How large must \(n\) be in order to ensure that with probability at least 0.9 the combined weight of the College \(B\) team exceeds that of the College \(A\) team by at least 50 lb?
A single stream of cars, each of width \(a\) and exactly in line, is passing along a straight road of breadth \(b\) with speed \(V\). The distance between the rear of each car and the front of the one behind it is \(c\). Show that, if a pedestrian is to cross the road in safety in a straight line making an angle \(\theta\) with the direction of the traffic, then his speed must be not less than \[\frac{Va}{c\sin\theta+a\cos\theta}.\] Show also that if he crosses the road in a straight line with the least possible uniform speed, he does so in time \[\frac{b}{V}\left(\frac{c}{a}+\frac{a}{c}\right).\]
A uniform beam of weight \(W\) stands with one end on a sheet of ice and the other end resting against the smooth vertical side of a heavy chair of weight \(\lambda W\). Show that the maximum inclination of the beam to the vertical is given by \(\tan^{-1} 2\mu\) or \(\tan^{-1} 2\lambda\mu\) according as the chair or the beam is the heavier, the coefficient of friction between the ice and beam, and the ice and chair, being \(\mu\).
A horizontal conveyor belt moves with a constant velocity \(u\). At time \(t = 0\), a parcel of mass \(m\) is dropped gently onto the belt. If the coefficient of friction between the parcel and the belt is \(\mu\), find
Two particles of masses \(m\) and \(2m\) are suspended over a movable pulley of mass \(m\) by a light string of length \(l\). The movable pulley is itself connected to a particle of mass \(4m\) by a light string of length \(L\) which passes over a fixed pulley. Find the acceleration of the particle of mass \(4m\). [You may neglect the moments of inertia of the pulleys.]
A uniform spherical dust cloud of mass \(M\) expands or contracts in such a way as to remain both uniform and spherical. The gravitational force on a particle of mass \(m\) at a distance \(r\) from the origin is radial and given by \[F = -\frac{4\pi}{3}G\rho mr,\] \(\rho\) being the density of the cloud and \(G\) the gravitational constant. By considering a particle on the surface of the cloud at distance \(R\) from the centre of the cloud, or otherwise, show that \[\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = -\frac{GM}{R_M},\] \(R_M\) being a constant. Verify that for \(R_M > 0\) this equation has a solution of the form \begin{align*} R &= a\sin^2\chi\\ t &= b(\chi-\sin\chi\cos\chi), \end{align*} where \(a\) and \(b\) are constants. Evaluate \(a\) and \(b\) in terms of \(G\), \(M\) and \(R_M\). Show that this solution describes a cloud that expands from infinite density at \(t = 0\) and which collapses back to infinite density at time \[t_\infty = \pi\sqrt{\frac{R_M^3}{2GM}}.\]
A spherical raindrop has mass \(m\), radius \(r\) and downward speed \(v\) as it falls through a cloud of water vapour, which is moving upwards at speed \(U\). The raindrop grows by the condensation of water vapour on its surface, so that the increase of mass per unit time is proportional to the surface area. The raindrop starts from rest with radius \(r_0\) at time \(t = 0\).