Ms. Bayes has a dog named Dorrie. If Ms. Bayes feeds Dorrie before going
to bed, Dorrie wakes Ms. Bayes before 5 a.m. only 10% of the the time. If Ms.
Bayes forgets to feed Dorrie before going to be bed, Dorrie wakes Ms. Bayes
before 5 a.m. 80% of the time. Ms. Bayes forgets to feed Dorrie one day in twenty.
If you hear Ms. Bayes complaining that Dorrie woke her before 5 a.m., what is
the probability that she forgot to feed Dorrie the previous evening?
Thanks for the help!
Originally Posted by ballin_sensation
P(forgot | wakes)=P(wakes | forgot)P(forgot)/P(wakes)
where P(wakes)=P(wakes | forgot)P(forgot)+P(wakes | remembered)P(remembered)