A three day festival is being planned to be held in a location with the following weather characteristics:
The probability of rain on any given day is 0.06.
If it rains on a particular day then the probability that it also rains on the next day is 0.37.
If it rains for two days in a row then the probability that it will rain on the following day is 0.21.
What is the probability that it will rain on any day during the festival?
Here is my solution:
Let $i$ denote the event it rains on the $i$th day.
$P(i)$ is the probability that it rains on the $i$th day.
$P(i^c)$ is the probability that it does not rain on the $i$th day
\begin{align*} P(1 \cup 2 \cup 3) &= P(1)+P(2)+P(3)-P(1 \cap2)-P(2 \cap 3)-P(1 \cap 3)+P(1 \cap 2 \cap 3) \\ &= 0.06+0.06+0.06-P(2|1)P(1)-P(3|2)P(2)-P(1 \cap 3)+P(3|1,2)P(2|1)P(1) \\ &= 0.18-0.37 \cdot 0.06-0.37 \cdot 0.06 - P(1 \cap 3) + 0.21 \cdot 0.37 \cdot 0.06 \\ &= 0.18-0.0222-0.0222-P(1 \cap 3)+0.004662 \\ &= 0.140262-P(1 \cap 3) \end{align*}
We couldn’t get $P(1 \cap 3)$ directly, but we could calculate $P(1 \cap 3)$ in terms of $P(3|1,2^c)$:
\begin{align*} P(1 \cap 3)&=P(1\cap2\cap3)+P(1\cap2^c \cap 3) \\&= P(3|1,2)P(2|1)P(1)+P(3|1,2^c)P(2^c|1)P(1) \\&=0.21\cdot 0.37\cdot0.06+P(3|1,2^c)\cdot(1-0.37)\cdot0.06 \\&=0.004662+0.0378\cdot P(3|1,2^c) \end{align*}
Since the lower bound (if the first day rains, the second day no rain, the third day always no rain) of $P(3|1,2^c)$ is $0$, and the upper bound (if first day rains, second day no rain, the third day always rain) is:
\begin{align*} max(P(3|1,2^c))&=P(3|2^c) \\&= P(3)-P(2 \cap 3)\\ &=P(3)-P(3|2)P(2)\\ &=0.06-0.37\cdot 0.06\\ &=0.0378 \end{align*}
so the upper bound of $P(1 \cap 3)$ is:
\begin{align*} max(P(1 \cap 3)) &=0.004662+0.0378\cdot P(3|1,2^c)\\ &=0.004662+0.0378\cdot0.0378\\ &=0.00609084 \end{align*}
Thus $P(1 \cap 3)$ is between $[0, 0.006]$, and the probability that it will rain on any day during the festival is between $[0.134,0.14]$