How do I add probabilities? Say I have 23% chance of A, and 48% chance of B, what are the chances of either? I used to think I would just add the probabilities, intuitively...then I came across problems where it sums to greater than 100%, but it’s not certain. If you think like I used to think, this abstract example won’t help you. So I’ll give a descriptive version below. For anyone who can explain it to me, feel free to skip the next part:
Say Jimmy wants to destroy an unwanted statue. From research on statue destruction, he believes there is a 95% chance that the statue will be destroyed if he places it in front of the train. He also knows from research that if he puts a rope around it and drops it from a height, the force of the snap back will break the statue in half. He reasons that he can put a rope around it then attach that rope to the tracks, so that when the train comes, the snapping force will apply with the same destructive probability that the research cites − 96%. If the statue isn’t destroyed, Jimmy is going to have to take care of the pieces, which will be a lot of work, so Jimmy wants to know what he’s getting into. I wonder if there a way to combine the probabilities for an overall probabiltiy given that both apply?
You take the probability of A not happening and multiply by the probability of B not happening. That gives you P(not A and not B). Then subtract that from 1. The probability of at least one of two events happening is just one minus the probability of neither happening.
In your example of 23% and 48%, the probability of getting at least one is
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
How do I add probabilities? Say I have 23% chance of A, and 48% chance of B, what are the chances of either? I used to think I would just add the probabilities, intuitively...then I came across problems where it sums to greater than 100%, but it’s not certain. If you think like I used to think, this abstract example won’t help you. So I’ll give a descriptive version below. For anyone who can explain it to me, feel free to skip the next part:
Say Jimmy wants to destroy an unwanted statue. From research on statue destruction, he believes there is a 95% chance that the statue will be destroyed if he places it in front of the train. He also knows from research that if he puts a rope around it and drops it from a height, the force of the snap back will break the statue in half. He reasons that he can put a rope around it then attach that rope to the tracks, so that when the train comes, the snapping force will apply with the same destructive probability that the research cites − 96%. If the statue isn’t destroyed, Jimmy is going to have to take care of the pieces, which will be a lot of work, so Jimmy wants to know what he’s getting into. I wonder if there a way to combine the probabilities for an overall probabiltiy given that both apply?
P(A or B) = P(A) + P(B) - P(A and B)
You use the Inclusion-Exclusion Principle.
You take the probability of A not happening and multiply by the probability of B not happening. That gives you P(not A and not B). Then subtract that from 1. The probability of at least one of two events happening is just one minus the probability of neither happening.
In your example of 23% and 48%, the probability of getting at least one is
1 - (1-0.23)*(1-0.48) = 0.60.
Only if A and B are independent.
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!