Hank's Gold Mine has a very long vertical shaft of height \(l\). A light chain of length \(l\) passes over a small smooth light fixed pulley at the top of the shaft. To one end of the chain is attached a bucket \(A\) of negligible mass and to the other a bucket \(B\) of mass \(m\). The system is used to raise ore from the mine as follows. When bucket \(A\) is at the top it is filled with mass \(2m\) of water and bucket \(B\) is filled with mass \(\lambda m\) of ore, where \(0<\lambda<1\). The buckets are then released, so that bucket \(A\) descends and bucket \(B\) ascends. When bucket \(B\) reaches the top both buckets are emptied and released, so that bucket \(B\) descends and bucket \(A\) ascends. The time to fill and empty the buckets is negligible. Find the time taken from the moment bucket \(A\) is released at the top until the first time it reaches the top again. This process goes on for a very long time. Show that, if the greatest amount of ore is to be raised in that time, then \(\lambda\) must satisfy the condition \(\mathrm{f}'(\lambda)=0\) where \[\mathrm{f}(\lambda)=\frac{\lambda(1-\lambda)^{1/2}} {(1-\lambda)^{1/2}+(3+\lambda)^{1/2}}.\]
Suppose that a solution \((X,Y,Z)\) of the equation \[X+Y+Z=20,\] with \(X\), \(Y\) and \(Z\) non-negative integers, is chosen at random (each such solution being equally likely). Are \(X\) and \(Y\) independent? Justify your answer. Show that the probability that \(X\) is divisible by \(5\) is \(5/21\). What is the probability that \(XYZ\) is divisible by 5?
Solution: They are not independent: \begin{align*} && \mathbb{P}(X = 20 \,\, \cap Y = 20) = 0 \\ && \mathbb{P}(X = 20 )\mathbb{P}(Y = 20) \neq 0 \\ \end{align*} \begin{align*} X = 0: && 21 \text{ solutions} \\ X = 5: && 16 \text{ solutions} \\ X = 10: && 11 \text{ solutions} \\ X = 15: && 6 \text{ solutions} \\ X = 20: && 1 \text{ solutions} \\ 5 \mid X: && 55 \text{ solutions} \\ \\ && \binom{20+2}{2} = 11 \cdot 21 \text{ total solutions} \\ \Rightarrow && \mathbb{P}(5 \mid X) = \frac{55}{11 \cdot 21} = \frac{5}{21} \end{align*} \begin{align*} \mathbb{P}(5 \mid XYZ) &= 3\cdot \mathbb{P}(5 \mid X) - 2\mathbb{P}(5 \mid X, Y, Z) \\ &= \frac{3 \cdot 55 - 2 \cdot \binom{4+2}{2}}{11 \cdot 21} = \frac{35}{77} \end{align*}
I have a bag initially containing \(r\) red fruit pastilles (my favourites) and \(b\) fruit pastilles of other colours. From time to time I shake the bag thoroughly and remove a pastille at random. (It may be assumed that all pastilles have an equal chance of being selected.) If the pastille is red I eat it but otherwise I replace it in the bag. After \(n\) such drawings, I find that I have only eaten one pastille. Show that the probability that I ate it on my last drawing is \[\frac{(r+b-1)^{n-1}}{(r+b)^{n}-(r+b-1)^{n}}.\]
To celebrate the opening of the financial year the finance minister of Genland flings a Slihing, a circular coin of radius \(a\) cm, where \(0 < a < 1\), onto a large board divided into squares by two sets of parallel lines 2 cm apart. If the coin does not cross any line, or if the coin covers an intersection, the tax on yaks remains unchanged. Otherwise the tax is doubled. Show that, in order to raise most tax, the value of \(a\) should be \[\left(1+{\displaystyle \frac{\pi}{4}}\right)^{-1}.\] If, indeed, \(a=\left(1+{\displaystyle \frac{\pi}{4}}\right)^{-1}\) and the tax on yaks is 1 Slihing per yak this year, show that its expected value after \(n\) years will have passed is \[ \left(\frac{8+\pi}{4+\pi}\right)^{n}.\]