What? If A=B, this would say that Pr(A)=1.Cognac said:Pr(A|B)=Pr(B|A)Pr(A)
No. A and B may be independent in general but not independent given Z. Example: toss two coins. A=coin 1 is heads. B=coin 2 is heads. Z=coins 1 and 2 match. Obviously Z forces a dependence between A and B.Also, if X&Y are independent, then would I get Pr(X&Y|Z)=Pr(X|Z)Pr(Y|Z)?
Thanks, that example about independence makes things make more sense now.FactChecker said:What? If A=B, this would say that Pr(A)=1.
No. A and B may be independent in general but not independent given Z. Example: toss two coins. A=coin 1 is heads. B=coin 2 is heads. Z=coins 1 and 2 match. Obviously Z forces a dependence between A and B.
By "divide by the conditional", I assume you mean "divide the right side by P(A)". Yes, Bayes' Rule always works. The only exception to the form of Bayes' Rule that you are using is if P(A) = 0. Then both P(B|A) and the division of the right side by 0 are problems. (Some statements of Bayes' Rule talk about a partitioning of the space. If you use ever that form, make sure that condition is met.)Cognac said:I forgot to divide by the conditional. But does Bayes Rule still work for the case I presented? Given two joint events, X&Y?
Conditional probability with joint condition is a mathematical concept that calculates the probability of an event A occurring, given that another event B has already occurred. It takes into account the probability of both A and B happening together.
Regular conditional probability only considers the probability of one event occurring, given that another event has occurred. Conditional probability with joint condition takes into account the probability of both events occurring together.
The formula for calculating conditional probability with joint condition is P(A|B) = P(A and B) / P(B), where P(A|B) represents the probability of event A occurring given event B has occurred, P(A and B) represents the probability of both events A and B occurring, and P(B) represents the probability of event B occurring.
Say you have a bag with 5 red marbles and 10 blue marbles. You randomly select two marbles from the bag without replacement. What is the probability of selecting a red marble on the second draw, given that the first marble drawn was red? In this scenario, the conditional probability with joint condition would be P(red on second draw | red on first draw) = 4/14 = 2/7.
Conditional probability with joint condition is used in various fields such as medicine, finance, and engineering to make predictions and decisions. For example, in medicine, it can be used to calculate the probability of a patient having a certain disease, given that they exhibit certain symptoms. In finance, it can be used to assess the risk of an investment, given that certain market conditions are present.