Let A represents a random event, and B represents another event. I understand that if A and B are mutually exclusive, then P(A n B) or the probabilities of A intersect with B will give 0.
What’s confusing is at times the lecture notes I have state that P(A n B) is equal to P(A) x P(B), but other times it gives a value for P(A n B), but it isn’t equal to that of P(A) x P(B). So when can I use P(A n B) = P(A) x P(B) to find the intersection of two events?
When A and B are independent events, or in other words when the probability of “A given B” is the same as the probability of A by itself.
Unfortunately, if you dig a little into the definition of conditional probability (i.e, what I mean when I say the probability of “A given B”) you’ll find that mathematically the statement P(A n B)=P(A) x P(B) is the definition of “A and B are independent”, not the other way around. But you can get a good grasp of the colloquial meaning of “independent”…namely, that the outcome of B has no effect on the outcome of A and vice versa…and you’ll get the idea.
For example, if A=“roll 6 on a die” and B=“draw a spade from a deck of cards”, the events are independent and P(A n B)=P(A) x P(B). But if A=“roll 6 on a die” and B=“roll an even number on that same die”, then they’re not independent and P(A n B) will not equal the product of the individual probabilities.
So what if P(A) and P(B) are dependent? Is there anyway to find P(A n B) if they A and B are dependent? I do know that if A is a subset of B, then P(A n B) = P(A). What if it is not? Is it possible to work out P(A n B) only with just P(A) and P(B) and if A and B are dependent and A is not a subet of B>
There is no general way. But you could to do it if you also knew P(A union B).
P(A intersection B) = P(A) + P(B) - P(A union B)