Sunny Days

Solution

Answer: 243.7

Let \(S_n\) denote the probability that day \(n\) is sunny. We are given the transition probabilities \[ \Pr(\text{sunny today} \mid \text{sunny yesterday}) = 0.7 \] \[ \Pr(\text{sunny today} \mid \text{rainy yesterday}) = 1 - 0.4 = 0.6 \] since if yesterday was rainy, the probability it is rainy today is \(0.4\). The first day is sunny, so \[ S_1 = 1. \] Day \(n+1\) can be sunny in two ways: • Yesterday was sunny and it stays sunny • Yesterday was rainy and it becomes sunny Thus \[ S_{n+1} = 0.7S_n + 0.6(1 - S_n). \] Simplifying, \[ S_{n+1} = 0.6 + 0.1S_n. \] To understand this sequence, find its equilibrium value \(S\): \[ S = 0.6 + 0.1S. \] So \[ 0.9S = 0.6 \] \[ S = \frac{2}{3}. \] Thus the probability approaches \( \frac{2}{3} \) in the long run. The recurrence solution is \[ S_n = \frac{2}{3} + \left(S_1 - \frac{2}{3}\right)(0.1)^{\,n-1}. \] Since \(S_1 = 1\), \[ S_n = \frac{2}{3} + \frac{1}{3}(0.1)^{\,n-1}. \] The expected number of sunny days in \(365\) days is \[ \sum_{n=1}^{365} S_n. \] Substitute the formula: \[ \sum_{n=1}^{365} \left( \frac{2}{3} + \frac{1}{3}(0.1)^{\,n-1} \right). \] Split the sum: \[ = 365 \cdot \frac{2}{3} + \frac{1}{3}\sum_{n=1}^{365}(0.1)^{\,n-1}. \] The geometric series is \[ \sum_{n=1}^{365}(0.1)^{\,n-1} = \sum_{k=0}^{364}(0.1)^k = \frac{1-(0.1)^{365}}{1-0.1}. \] So \[ = \frac{1-(0.1)^{365}}{0.9}. \] Therefore \[ \text{Expected sunny days} = 365\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1-(0.1)^{365}}{0.9}. \] Since \[ \frac{1}{3} \cdot \frac{1}{0.9} = \frac{10}{27}, \] this becomes \[ \frac{730}{3} + \frac{10}{27}\left(1-(0.1)^{365}\right). \] Because \( (0.1)^{365} \) is astronomically small, it is essentially zero. \[ \frac{730}{3} + \frac{10}{27} \approx 243.7037. \] Thus the expected number of sunny days in the year is approximately 243.7.