Question:
90% likelihood to meet her parents will happen tonight and it will be only held at three places, (C), (N), (S) with the same probability. You have been to (C) and (N) but you haven't found the parents. What is the probability that you see the parents at (S)?
From the question since there is a 90% likelihood to meet there is a 10% not to meet
I know I need to use Bayes theorem:
and trying to find the $P(\text{seeing parents at (S)}|\text{parents not at (C) and (S)})$
$P(\text{seeing parents at (S)} = 0.3$ from the question
but i am unsure how to proceed to find
$P(\text{parents not at (C) and (S)})$
and
$P(\text{parents not at (C) and (S)})|P(\text{seeing parents at (S)})$
to finish the bayes theorem
$$ P( S | \text{not C and not N})=\frac{P(S) P(\text{not C or N} | S)}{P(S) P(\text{not C or N} | S) + P(\text{not S} | \text{not C or N} )} $$
For $P(S)$, you also have to multiply by that $0.9$ probability of the parents actually arriving. $$P(S) = 0.9 \times 0.3 $$ Clearly, $$P(\text{not C or N}| S) =1 $$ and $$P(\text{not S}|\text{not C or N} )=\frac{P(\text{not S,C,N })}{P(\text{not C or N})} =\frac{0.1}{0.9\cdot 0.3 +0.1}$$