How can I reword this problem illustrating a scenario that needs Bayes Theorem to solve? Taken from Stat Trek, an example explaining Bayes Theorm http://stattrek.com/probability/bayes-theorem.aspx

Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on the day of Marie's wedding?

Can someone help reword this problem?. I don't particular understand what it means "When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time".
How can I assign the variables to fit this formula?

 A: Let the probability that it actually rains be $R$. Our prior information about $R$ is based on the "past few years" and it tells us that it rains, on average, 5 days a year (let's say three non-leap years just to finesse the problem of leap days) so we know that:
$$
P(R) = \frac{5}{365} = \frac{1}{73} \approx .0137
$$
Now let's talk about the weather forecaster's accuracy (or lack thereof). Let $P$ represent the probability that Mr. Sunshine accurately predicts rain. We know that his true positive rate (accurate prediction of presence of event) is 90%, so we know $P(P|R)$, the probability that his prediction of rain occurs when (given) it actually rains. We also know his false positive rate, the probability that the test (his prediction) indicates true when the phenomenon (rain) is false, $P(P|\overline{R})$ is 10%. 
Lastly, poor Marie is in possession of a $P$, a prediction of rain. She wants to know $P(R|P)$, the probability that it will rain given the weather report. Now we can use Bayes Theorem:
$$
\begin{align}
P(R|P) &= \frac{P(R \cap P)}{P(P)}\\
P(R \cap P) &= P(P \cap R) = P(P|R)P(R)\\
P(P) &= P(P \cap R) + P(P \cap \overline{R})\\
P(P \cap \overline{R}) &= P(P|\overline{R})P(\overline{R})
\end{align}
$$
So, fully spelled out in terms of the data we have, we need:
$$
P(R|P) = \frac{P(P|R)P(R)}{P(P|R)P(R) + P(P|\overline{R})P(\overline{R})}
$$
Which when we substitute values becomes
$$
\begin{align}
P(R|P) &=\frac{\frac{9}{10}\times \frac{1}{73}}{\left(\frac{9}{10}\times\frac{1}{73}\right)+\left(\frac{1}{10}\times \frac{72}{73}\right)}\\
&=\frac{\frac{9}{730}}{\frac{9}{730} + \frac{72}{730}}\\
&=\frac{9}{81} = \frac{1}{9} \approx 11.1\%
\end{align}
$$
A: We have only 4 possibilities to consider:
1) The probability that he predicts rain AND it does rain = $(0.90)(\frac{5}{365})$
2) The probability that he predicts rain AND it doesn't rain = $(0.10)(\frac{360}{365})$
3) The probability that he predicts no rain AND it does rain = $(0.10)(\frac{5}{365})$
4) The probability that he predicts no rain AND it doesn't rain = $(0.90)(\frac{360}{365})$
We are given the valuable piece of information that rain has already been predicted. Therefore, the only possible "world" we need to be concerned with is the world where rain is predicted. Items #1 and #2 are the only components of our world. Everything else can be excluded.
We are left with;
The probability that it will rain = $$\frac{(0.90)(\frac{5}{365})}{(0.90)(\frac{5}{365})+(0.10)(\frac{360}{365})}$$
or $11.1$%
Here is the fill-in:
$$P(Rain|RainPred) = \frac{P(RainPred|Rain)P(Rain)}{P(RainPred|Rain)P(Rain)+P(RainPred|NoRain)P(NoRain)}$$
This is the way I read it out loud:
The probability of "rain when rain has been predicted" is equal to the probability of "rain has been predicted when it rains" times the probability of "rain" divided by the probability of "rain has been predicted when it rains" times the probability of "rain" plus the probability of "rain has been predicted when it doesn't rain" times the probability of "not raining".
Converting probability problems to plain but consistent English is the quickest way to achieving understanding of the problem. Lots of times it seems the problem statements are written like riddles. The textbook I had in college was lousy so that didn't help at all. One needs to get a clear picture of what is going on in order to grasp probability problems. For me, probability mathematics is very hard to grasp because many times it's very difficult to form a mental picture of what is going on. Probability math, in my case, seems to require much more work and time to achieve understanding.
