Becky has four umbrellas. During her workday, she travels between home and the office. She takes an umbrella only when it rains; otherwise, it remains where it was last—in the office or at home. Suppose on a given day that all her umbrellas are in the office, whereas she’s at home preparing for the office, and if it rains, she will get wet. The question is:
If the location has a 60% probability of rain, what is the chance that Becky gets wet?
The problem can be solved as a Markovian chain. For that, we must divide the conditions into five states. They are
0: no umbrella state
1: one umbrella
2: two umbrellas
3: three umbrellas
4: four umbrellas
We must know what movements are possible from one state to another to develop translation probabilities.
From 0: Becky must go from 0 to 4, from one place without an umbrella to the other with all umbrellas. As this must happen irrespective of whether it rains or not, the probability of this movement is 1.
From 1: If it rains, p = 0.6, Becky carries the umbrella with her. In other words, she goes from state 1 to state 4 (3 already + 1 incoming).
If it doesn’t rain, p = 0.4, she will go from state 1 to state 3.
From 2: If it rains, state 2 to state 3. If it doesn’t rain, she will go from 2 to 2.
From 3: If it rains, 3 to 2; if it doesn’t, 3 to 1.
From 4: If it rains, 4 to 1; if it doesn’t, 4 to 0.
Here is the diagram representing the chain.
The translation matrix is
The required task is to find the stable end distribution of cities, which can be done using the relationship.
Xn = Pn X0
We use the Matrix calculator for P100
Multiply this with any starting distribution, we get the end state probabilities as,
The probability that she’s at state 0, P(0) = 0.09. Since the probability of rain when Becky is at 0 state is 0.6, the chance she gets drenched is 0.09 x 0.6 = 0.054 or about 5%.
