I am trying out the first probability problem at this link whose solution is here.
The problem and the question of concern are as follows:
The solution given is:
If I understand correctly, they are reasoning that:
P(rain | saw forecast) = P(rain | missed forecast).
This can only happen if P(rain) is independent of P(saw/missed forecast):
P(rain) = P(rain | saw forecast) = P(rain | missed forecast).
This hinges on the detail initially given, that P(rain/not rain) is only dependent on P(forecast rain / forecast not rain), which is the part boxed in red below:
This assumption allows the conditional prbabilities P(rain) and P(not rain) on the two branches (saw / missed forecast) to be equal when the tree is expanded to include these probabilities.
However, P(saw forecast) and P(umbrella) are dependent on one another, that is why they have to back calculate P(saw forecast | umbrella and not raining) after figuring out the conditional probabilities in the upper branch (missed forecast).
This is the only way all of that makes sense to me. Can someone confirm?
From the upper part of the tree, we can immediately figure that
P(actually rain|missed forecast) = $0.1+0.7 p$
From the lower part of the tree, we realize that
P(actually rain|saw forecast) = $0.8p+0.1(1- p)=0.1+0.7p$
Hence, you are right that
P(actually rain)=P(actually rain|missed forecast)=P(actually rain|saw forecast)
which makes intuitive sense.
At the same time, you are also right that
P(actually rain|saw forecast)=
= P(actually rain|saw forecast, forecast rain)P(forecast rain|saw forecast) + P(actually rain|saw forecast, forecast not rain)P(forecast not rain|saw forecast)=
= $0.8p+0.1(1-p)$