91Ó°ÊÓ

First, we will find the probability of having \(k=0,1,..,5\) rainfalls within a 5-day period using the Poisson distribution formula: \(P(X=k) = \frac{(\lambda T)^k e^{-(\lambda T)}}{k!}\) where \(X\) represents the number of rainfalls, \(\lambda=0.2\) is the rate per day, \(T=5\) is the number of days, and \(k\) is the number of rainfalls we want to compute.

Next, we will calculate the probability of the water level being empty with at least \(k=0,1,...,5\) rainfalls. The total depletion in 5 days is 5,000 units per day which is 25,000 units. The possible water refilled is 5,000 units with probability 0.8 and 8,000 units with probability 0.2. Let \(E_k\) denote the event of having the reservoir empty with \(k\) rainfalls. For each \(k\), we can compute the probability \(P(E_k)\) by considering all possible combinations of 5,000 units and 8,000 units amounts of water refilled. Afterward, we find the probability of the reservoir being empty as: \(P(\text{Reservoir empty after 5 days}) = \sum_{k=0}^{5} P(E_k | X=k) P(X=k)\)

Now, to calculate the probability of the reservoir being empty within the next 10 days (sometime in the first 10 days, not exactly on the tenth day), we will consider all possible scenarios. We will iterate through the days and compute the probability of the reservoir being empty on each day. For each day \(t=1,2,\dots,10\), we will first find the number of rainfalls until day \(t\), denoted by \(X_t\). Then, we will calculate the probability of the reservoir being empty on the day \(t\), \(P(E_{X_t})\), using the method in part (a). Finally, we will find the probability of the reservoir being empty sometime within the next 10 days as: \(P(\text{Reservoir empty within 10 days}) = 1 - P(\text{Reservoir not empty any day within 10 days}) = 1 - \prod_{t=1}^{10} (1-P(E_{X_t}))\)