Physical disasters and probability An urban area is susceptible to one earthquake per year with probability $30 \%$ and two with probability $5 \%$, also three earthquakes per year is impossible.
The same area is also susceptible to floods, which may result from heavy rain (a fact which occurs with probability $100 \%$ within a year) or from a Dam failure due to the earthquake (event occurring with probability $25 \%$ in a year). The two events (floods from rain and floods from dam damage) are considered to be independent.
Calculate the probability of a flood in a year (for the specific area).
I know it can be solved by using Bayes theorem but am not sure on how to use it.
$P(\text{heavy_rain}) = 1$
$P(\text{dam_failure_from_earthquake}) = 0.25$
$P(\text{1_earthquake}) = 0.3$
$P(\text{2_earthquakes}) = 0.05$
$P(\text{3_earthquakes}) = 0.00$
$P(\text{floods}) = P(\text{floods_from_rain})+P(\text{floods_from_dam})$
$P(\text{floods_from_rain}) = 1$
but I don't know how to continue.
 A: According to your description,
$$\mathbb{P}(\text{flood}) = \mathbb{P}(\text{flood} | \text{heavy rain}) \mathbb{P}(\text{heavy rain}) = 1 \cdot 1 = 1.$$
Clearly, $\mathbb{P}(\text{flood} | \text{heavy rain}) \neq 1$ or $\mathbb{P}(\text{heavy rain}) \neq 1$, otherwise the problem is trivial.
Assuming that you transcribed the problem incorrectly and

*

*$\mathbb{P}(\text{flood} | \text{heavy rain}) = x\in [0,1]$

*$\mathbb{P}(\text{heavy rain}) = y\in [0,1]$,

then
\begin{align*}
\mathbb{P}(\text{flood})
&= \mathbb{P}(\text{flood} | \text{no earthquake, heavy rain}) \cdot \mathbb{P}(\text{no earthquake, heavy rain}) \\
&\quad+ \mathbb{P}(\text{flood} | \text{1 earthquake, heavy rain}) \cdot \mathbb{P}(\text{1 earthquake, heavy rain}) \\
&\quad+ \mathbb{P}(\text{flood} | \text{2 earthquakes, heavy rain}) \cdot \mathbb{P}(\text{2 earthquakes, heavy rain}) \\
&\quad+ \mathbb{P}(\text{flood} | \text{no earthquake, no heavy rain}) \cdot \mathbb{P}(\text{no earthquake, no heavy rain}) \\
&\quad+ \mathbb{P}(\text{flood} | \text{1 earthquake, no heavy rain}) \cdot \mathbb{P}(\text{1 earthquake, no heavy rain}) \\
&\quad+ \mathbb{P}(\text{flood} | \text{2 earthquakes, no heavy rain}) \cdot \mathbb{P}(\text{2 earthquakes, no heavy rain}) \\
&= x \cdot 0.65 y \\
&\quad+ \big[0.25 + x - 0.25 x\big] \cdot 0.30 y \\
&\quad+ \big[(1 - 0.75^2) + x - (1 - 0.75^2)x\big] \cdot 0.05 y \\
&\quad+ 0 \cdot 0.65 (1 - y) \\
&\quad+ 0.25 \cdot 0.30 (1 - y) \\
&\quad+ (1 - 0.75^2) \cdot 0.05 (1 - y).
\end{align*}
This is the completely general answer.
Just replace $x$ and $y$ by the values given in your (corrected) statement (i.e. not $x = y = 1$).
Note : the quantities inside the two brackets are calculated by the inclusion-exclusion principle.
