Search results
Results From The WOW.Com Content Network
So then is it true that if you take the complement of P(A U B) which is .79 and add to P(A) which is .15 you get .94 the same way??? Or am I just making stuff up haha? $\endgroup$ – jc707270
Then, the probability of only A occurring is the probability of A occurring given that only one of the events will occur, or P(A ∣ S), where S is the event that only one of A and B occurs. Then the answer is P (A ∩ S) P (S) = P (A) P (A ∪ B) − P (A ∩ B) =.75.8 =.9375. This doesn't seem correct or simple enough. Any advice is appreciated.
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
union of two independent probabilistic event. I have following question: Suppose we have two independent events whose probability are the following: P(A) = 0.4 and P(B) = 0.7. We are asked to find P(A ∩ B) from probability theory. I know that P(A ∪ B) = P(A) + P(B) − P(A ∩ B). But surely the last one is equal zero so it means that ...
The webpage discusses calculating the conditional probability of three events occurring simultaneously.
First, you probably need to switch around the probabilities on the right. If a <b a <b, then P(x <a) ≤ P(x <b) P (x <a) ≤ P (x <b), so the right-hand side of your expression could be negative. It's probably more intuitive if you rewrite it as. P(x <b) = P(x <a) + P(a <x <b) P (x <b) = P (x <a) + P (a <x <b)
I'm having some difficulty understanding Bayes' theorem with multiple events. I'm trying to put together a Bayesian network. I have four independent probabilities but I have found that A, B and C ...
I think I would do (4/7) * (3/6). P(A and B) is found by multiplying the two. The key thing here is that you are getting the probability $\frac36$ by assuming there are only $6$ marbles remaining and $3$ of them are yellow.
One of the property of Independent events is that the probability of their intersection is a product of their individual probabilities. So, P(A ∩ B) P (A ∩ B) is P(A) × P(B) P (A) × P (B). Whereas for mutually exclusive events, the probability of intersection is 0 0 as they can't both occur simultaneously! P(A ∪ B ∪ C) = P(A) + P(B ...
The condition P(A ∣ B) = P(A) means that knowing that B happened has no influence on the probability of A, which is the typical Bayesian view on the matter. The probability is the same with or without knowledge of B. In contrast, your definition says that whether B happens or not gives the same conditional probablity of A.