Conditional Probability vs Dependent Events I'm conceptually/intuitively confused about when we would use the joint probability (multiplication rule) versus a conditional probability, specifically when the dependent event has no chance of occurring without the other event.
Here's an example that illustrates my confusion:
Suppose we have a weird city, where there's a 20 percent chance of rain (so a 80% chance of any other weather). Now suppose that it can only rain frogs when it rains, and there's a 10% chance of frogs.
Why can we not use bayes rule to model:
$$Pr(frogs | rain) = \frac{Pr(rain | frogs)\cdot Pr(frogs)}{Pr(rain)} = \frac{1 \cdot 0.1}{0.2}$$
I know the correct way of doing it would be the multiplication rule, but why can we not use a conditional probability? We know that the probability of raining frogs is 0.1 because it cannot occur in any other scenario other than when it rains, and if there are raining frogs, we must have rain so $Pr(rain | frogs)=1$. And we also have $Pr(rain)=0.2$, so it seems like we have everything we need for conditional probability.
The correct answer would of course be $0.2\cdot0.1=0.02$.
What am I missing aside from the nonsensical results?
Edited for sensical numbers, but assume that the probability of raining frogs is just the proportion of days on which it rains.
 A: 
"We know that the probability of raining frogs is 0.1 because it cannot occur in any other scenario other than when it rains"

This is false. $P(frogs|rain)=.1$ but $P(frogs)\ne.1$!
So it is incorrect to plug it into Bayes' Theorem as you did.

"The correct answer would of course be 0.2⋅0.1=0.02."

You found P(frog), not P(frog|rain). Indeed,
$\begin{split}P(frog)&=P(frog|no rain)P(no rain)+P(frog|rain)P(rain)\\
&=0+.1*.2\\
&=.02\end{split}$
Conclusion
You switched P(frogs|rain) and P(frogs) around. One is the conditional probability of frogs given that it rains and one is the overall probability that it rains frogs.
A: You have your definitions all mixed up. You have already stated that $P(\text{frogs}|\text{rain})=0.1$, so you don't need Bayes Theorem to prove it.
What you need is the law of total probability, which states that:
$$P(\text{frogs})=P(\text{frogs}|\text{rain})P(\text{rain}) + P(\text{frogs}|\text{no rain})P(\text{no rain})$$
which gives
$$P(\text{frogs})=(0.1\times0.2) + (0.0\times0.8)=0.02$$
