The other responses are absolutely correct. If they clarified things for you, great!
However, you indicated that your intuition wasn’t getting you to the correct answer, so let’s try to formulate these into a new intuition.
You are starting along a path from left to right. The left side is where you started, and way over on the right side are different outcomes.
The ‘chances’ you mention above are forks in the path, and you are taking each path with some probability.
At the first fork there are two paths, one in which A happened and one in which A did not. The path where A happened is picked 23% of the time.
These two paths (A and not A) each branch again at the second fork, one in which B happened and one in which B did not. At this fork, the path where B happened is picked 48% of the time.
Two forkings means there is a total of 4 end points: One for (A, B), one for (A, not B), one for (not A, B), and one for (not A, not B).
The chance of ending up in each endpoint is equal to the product of chances.
If A happens with 23% chance and B happens with 48% chance, then both happening is (0.23 times 0.48).
The chance of something not happening is (1.00-(chance of happening)).
So, (A, not B) is (0.23 times (1.00-0.48))
(not A, B) is ((1.00-0.23) times 0.48)
and lastly: (not A, not B) is ((1.00-0.23) times (1.00-0.48))
When you say you want either, you are saying that you consider the (not A, not B) endpoint unacceptable, but that you are satisfied with the other three: (A, B), (A, not B), and (not A, B) are all equally good.
RolfAndreassen is showing that you can find the probability of either A or B by taking a double negative: since the chance of something not happening is (1.00-(chance of happening)), we can take (1.00-(chance I end up in the single bad endpoint)) which is equal to (1.00-(chance I end up at (not A, not B)) = (1.00 - ((1.00-0.23) times (1.00-0.48)))
Sarunas is showing that naively adding the probabilities of A and B returns too large a value. Only by subtracting out the probability of (A, B) can it be corrected.
Intuiting the probabilities as branching paths I found very helpful, especially when you get to situations where the probabilities depend on eachother. See http://www.yudkowsky.net/rational/bayes/ for an example
The other responses are absolutely correct. If they clarified things for you, great! However, you indicated that your intuition wasn’t getting you to the correct answer, so let’s try to formulate these into a new intuition.
You are starting along a path from left to right. The left side is where you started, and way over on the right side are different outcomes. The ‘chances’ you mention above are forks in the path, and you are taking each path with some probability.
At the first fork there are two paths, one in which A happened and one in which A did not. The path where A happened is picked 23% of the time. These two paths (A and not A) each branch again at the second fork, one in which B happened and one in which B did not. At this fork, the path where B happened is picked 48% of the time.
Two forkings means there is a total of 4 end points: One for (A, B), one for (A, not B), one for (not A, B), and one for (not A, not B).
The chance of ending up in each endpoint is equal to the product of chances. If A happens with 23% chance and B happens with 48% chance, then both happening is (0.23 times 0.48). The chance of something not happening is (1.00-(chance of happening)). So, (A, not B) is (0.23 times (1.00-0.48)) (not A, B) is ((1.00-0.23) times 0.48) and lastly: (not A, not B) is ((1.00-0.23) times (1.00-0.48))
When you say you want either, you are saying that you consider the (not A, not B) endpoint unacceptable, but that you are satisfied with the other three: (A, B), (A, not B), and (not A, B) are all equally good.
RolfAndreassen is showing that you can find the probability of either A or B by taking a double negative: since the chance of something not happening is (1.00-(chance of happening)), we can take (1.00-(chance I end up in the single bad endpoint)) which is equal to (1.00-(chance I end up at (not A, not B)) = (1.00 - ((1.00-0.23) times (1.00-0.48)))
Sarunas is showing that naively adding the probabilities of A and B returns too large a value. Only by subtracting out the probability of (A, B) can it be corrected.
Intuiting the probabilities as branching paths I found very helpful, especially when you get to situations where the probabilities depend on eachother. See http://www.yudkowsky.net/rational/bayes/ for an example
Hope this helps!