Each time it rains over the Cabbibo dam, a volume \(V\) of water is deposited, almost instanetaneously, in the reservoir. Each day (midnight to midnight) water flows from the reservoir at a constant rate \(u\) units of volume per day. An engineer, if present, may choose to alter the value of \(u\) at any midnight.
Suppose that it rains at most once in any day, that there is a probability \(p\) that it will rain on any given day and that, if it does, the rain is equally likely to fall at any time in the 24 hours (i.e. the time at which the rain falls is a random variable uniform on the interval \([0,24]\)). The engineers decides to take two days' holiday starting at midnight. If at this time the volume of water in the reservoir is \(V\) below the top of the dam, find an expression for \(u\) such that the probability of overflow in the two days is \(Q\), where \(Q < p^{2}.\)
For the engineer's summer holidays, which last 18 days, the reservoir is drained to a volume \(kV\) below the top of the dam and the rate of outflow \(u\) is set to zero. The engineer wants to drain off as little as possible, consistent with the requirement that the probability that the dam will overflow is less than \(\frac{1}{10}.\) In the case \(p=\frac{1}{3},\) find by means of a suitable approximation the required value of \(k\).
Suppose instead that it may rain at most once before noon and at most once after noon each day, that the probability of rain in any given half-day is \(\frac{1}{6}\) and that it is equally likely to rain at any time in each half-day. Is the required value of \(k\) lower or higher?
Solution:
It cannot overflow on the first day, since it is already \(V\) below the top. The only way it can overflow is if it rains both days. This will occur with probability \(p^2\). The probability it overflows therefore is the probability that bad timing hampers us, ie \(V - u(1+t_2) > 0\) where \(t_2\) is the timing of the rain on day 2 (as a fraction of a day). Ie \(t_2 < \frac{V}{u}-1\). Therefore
\begin{align*}
&& Q &= p^2 \left (\frac{V}{u} - 1 \right) \\
\Rightarrow && u &= \frac{Vp^2}{p^2+Q}
\end{align*}
The probability the reservoir overflows during this \(18\) days is \(\mathbb{P}(\text{rains more than }k\text{ times})\). The number of times it rains (\(X\)) is \(B(18, \tfrac13)\), since \(18 \cdot \tfrac13 = 6 > 5\) a normal approximation is reasonable, ie \(X \approx N(6, 4)\).
We wish to find \(k\) such that \(\mathbb{P}( X > k + 0.5) < \tfrac1{10}\) therefore \(k \approx 1.28 \cdot 2 + 6 - 0.5 \approx 8.1\) so they should set \(k\) to \(9\)
In this case we have \(B(36, \tfrac16)\) approximated by \(B(6, 5)\) which has a larger standard deviation, therefore we need to choose a larger value for \(k\).
[It turns out to actually be the same, but there's no reason to be able to expect students without a calculator to establish this]