This question is based on a previous question that I asked here. Same context as before:
Tom is a night owl and is very hungry at night. On day , he eats at home with probability 1− (0<<1); or, with probability , he goes out and goes the -th restaurant in his city. At any restaurant that he goes to, either:
(i) the restaurant is open with probability (where 0<<1). In that case, Tom orders some food for take-away;
(ii) the restaurant is closed so Tom returns home and has sleep for dinner because he is sad.
We assume that the collection of all events of the form {Tom stays home on day } and {The -th restaurant is open on day }, for =1,2,…, are (mutually) independent.
I stated that the conditional probability of Tom staying home on day 1, given that he did not order food on that day is
$$\frac{1-p}{p(1-q)}$$
However, I think that previous probability is wrong, and that instead it should actually be
$$\frac{1-p}{p\left(1-q\right)+(1-p)}=\frac{1-p}{1-pq}$$
Because either Tom goes out and the restaurant was closed with probability of (1−) meaning that he didn't order food and Tom stays at home with probability of 1− which also means that he didn't order food.
Could someone please tell me which one of these two conditional probabilities is correct?
I want to believe it's the second one. Thanks for any feedback.
Whenever confused re Bayes' Rule, it is a good idea to draw a contingency diagram
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ Went Out $\;\;\;\;$ Stayed home
Ordered food $\;\;\;\;\;\;\;\;$pq $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0$
Didn't order food $\;\;\;$p(1-q) $\;\;\;\;\;\;\;\;$ (1-p)
You should be able to clearly see that your second formula is correct