I have a question which I'm not sure how to phrase hence i couldn't find similar questions:
Usually all the examples i find online refer to dependent events and their probability of happening based on the event they depend on.
But in my question the issue is finding the probability of that event they depend on if i know that the dependent events happened and i know their own probability.
Example:
The chance if it raining today is 12%
when it rains, there is a 60% chance of sheep roaming the fields
on a not rainy day this probability is 70%
when it rains, there is 14% chance of new flowers blooming
on a not rainy day this probability is 24%
when it rains, there is a 1% chance of an alien invasion
- on a not rainy day this probability is 11%
Now if i know that today sheep roamed the fields, new flowers bloomed, and aliens invaded, what is the chance that it rained today? is it still 12% or does the probability chance because of the dependent events?
To elaborate on the comments:
First of all, to proceed we need to assume something about how the events depend on each other. It can't be the case that they are strictly independent of each other. For example: seeing the aliens is strong evidence for the absence of rain, and as the absence of rain also increases the probability of the other two events, seeing aliens increases our estimate for their probability as well. Best we can do (and it suffices here) is to assume that the three events are independent conditioned on the presence or absence or rain. That is, if we are told that it is raining then seeing aliens tells us nothing further about the sheep or flowers. Same if we are told it is not raining. Granted, this is a somewhat technical point but these things matter.
Under that assumption:
Let $\{\phi_A,\phi_B,\phi_C\}$ be the the probabilities of the three events when it is raining. Let $\{\psi_A,\psi_B,\psi_C\}$ be their probabilities when it is not. Let $P$ be the probability that it is raining. Then, by Bayes' Theorem, the probability that it is raining given that we observe all three events is $$\frac {P\times \phi_A\times \phi_B\times \phi_C}{P\times \phi_A\times \phi_B\times \phi_C+(1-P)\times \psi_A\times \psi_B\times \psi_C}$$
In your specific example that comes to $$\frac {.12\times .6\times .14\times .01}{.12\times .6\times .14\times .01+.88\times .7\times .24\times .11}=0.006160164$$
So we can be reasonably sure that it is not raining (small comfort, what with the aliens and all).