Conditional probabilities word problem ($A|B$) > Seamus the dog howls at the moon. The fuller the moon the more likely it is that Seamus will howl at it. Before the moon reaches about 90% fullness the probability Seam...--prophetes.ai

Conditional probabilities word problem ($A|B$) > Seamus the dog howls at the moon. The fuller the moon the more likely it is that Seamus will howl at it. Before the moon reaches about 90% fullness the probability Seamus will howl at it is 35%. Once the moon if 90% or more full, there is a 95% chance that Seamus will howl. You hear Seamus howl, what's the probability that the moon is not yet 90% full? I understand that we can break this up into conditional probabilities, with $A=\textrm{Moon is 90% full}$ and $B = \textrm{Seamus will howl}$. So what we are trying to find is $P(\neg A|B)$. So, of course, we know $P(B|A)=0.95$ and $P(B|\neg A)=0.35$. But how do we reverse this conditional probability? Thanks

Let $M$ be the event of the Moon being full $90\%$ or more and $H$ be the event of the dog howling. It is given: $$P(M)=0.1,P(M')=1-P(M)=0.9;\\\ P(H|M)=0.95, P(H'|M)=1-P(H|M)=0.05;\\\ P(H|M')=0.35, P(H'|M')=1-P(H|M')=0.65;\\\ P(M|H)=?\\\$$ The Bayes' rule (conditional probability): $$ P(M|H)=\frac{P(M\cap H)}{P(H)}=\frac{P(M)\cdot P(H|M)}{P(M)\cdot P(H|M)+P(M')\cdot P(H|M')}$$ Can you finish?