I've been searching through posts on this web about how to apply Bayes' rule on the question on my assignment. I've came across many useful tips and advices. However, not all of my questions are answered.
So I was trying to figure out how to calculate probability using Bayes rule on multiple conditions like P(a,b | c). I came across this posts https://math.stackexchange.com/q/408833 which explained how to solve this very nicely.
The question on my assignment :
So let ho = Hangover, f = flu, s = sore throat, n = nauseous and hd = headache.
Given that P(ho | hd) = 0.5
P(ho | n) = 0.5
P(ho | s) = 0.1
P(f | hd ) = 0.4
P(f | n) = 0.2
P(f | s) = 0.6
P(ho) = 0.3
P(f) = 0.05
and if there is nothing wrong with John (ie. no hang over or flu) the prior probability is 0.65. Flu and hangover cannot happen at the same time.
I applied the rule I found on my question, so I got :
P(ho, f|s) = P(ho | f,s)P(f|s)
my question didn't give the value for P(f,s) so I tried to use the fact that P(f,s) is
P(f AND s) = P(f | s) P(s)
This is where I encountered a second problem. The question didn't state the value of P(s) either. How do I calculate for a single P(a) if it is not given?
Another question I wanted to ask is that, I have also stumbled upon this posts Is $P( A \cup B\, |\, C) $ the same as $P(A | C) + P(B | C) $ ( $A$ and $B$ are mutually exclusive) they say that P( a U b | c ) is equal to P(a |c) + P(b|c).
In my assignment, it said that hangover and flu cannot happent at the same time. So does that mean P(ho,f|s) is equal to P(ho U f | s)? if so, Can I say that P(ho,f|s) = P(ho | s ) + P(f |s) ?
Sorry for writing and asking so many things here but I really need to get these question answered. Thank you in advance!