Conditional Probabilities Complementary Proof

[rbzima]

I'm having trouble seeing how this works out. It's blatantly obvious that this is true, but somehow I can't seem to get anywhere on paper with it to simplify it down to anything. Any help would be greatly appreciated!

$$P\left(A\right|B)=1-P\left(not A\right|B)$$

 

[EnumaElish]

P(A|B) = P(A&B)/P(B)
P(~A|B) = P(~A&B)/P(B)

What is P(A&B) + P(~A&B) = ?

 

[tacman]

Conditional probabilities can often be confusing to understand, so it's completely normal to have trouble with it. Let's break down the equation to better understand it.

P(A|B) represents the probability of event A occurring given that event B has already happened. This can also be written as "the probability of A given B."

On the other hand, P(not A|B) represents the probability of event A not occurring given that event B has already happened. This can also be written as "the probability of not A given B."

Now, the complement of an event is the opposite of that event. In this case, the complement of A would be not A, and the complement of not A would be A.

So, the equation P(A|B)=1-P(not A|B) is saying that the probability of event A occurring given B, is equal to 1 minus the probability of event A not occurring given B.

This makes sense because the probability of all possible outcomes must add up to 1. So, if the probability of event A occurring given B is subtracted from 1, it leaves the probability of event A not occurring given B.

I hope this helps to simplify the concept for you. Remember, practice makes perfect, so keep working on problems and asking for help when needed. Good luck!