I Conditional and joint probabilities of statistically dependent events

AI Thread Summary
In the discussion about conditional and joint probabilities of statistically dependent events, it is highlighted that both types of probabilities are interdependent. Conditional probability can be calculated if joint probability is known, and vice versa. However, without knowledge of either probability, one cannot determine the independence of the events. The confusion arises from the assumption that both probabilities must be calculated simultaneously. Understanding the relationship between these probabilities is crucial for accurate statistical analysis.
PainterGuy
Messages
938
Reaction score
73
Hi,

If the events A and B are statically dependent then the following formulas are used to calculate conditional probability and joint probability but there is a problem. As I see it both formulas are dependent upon each other. One cannot calculate conditional probability without first calculating joint probability, and one cannot calculate joint probability without knowing condition probability! Where am I going wrong? Could you please help me? Thank you!
1631537245645.png
 
Physics news on Phys.org
PainterGuy said:
As I see it both formulas are dependent upon each other. One cannot calculate conditional probability without first calculating joint probability, and one cannot calculate joint probability without knowing condition probability! Where am I going wrong? Could you please help me?
Well, you have to know something. If you know the conditional probability then you can calculate the joint probability. If you know the joint probability then you can calculate the conditional probability. If you don't know either then you cannot even tell if they are independent or not.
 
Reply
  • Like
Likes pbuk, PainterGuy and FactChecker

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
View full post »

Similar threads

B Decision for conditional probability instead of intersection of events
Replies
3
Views
2K
B The Probability of an Event
Replies
11
Views
2K
A A game of seemingly mutually independent events
Replies
7
Views
2K
I Linear regression and random variables
Replies
30
Views
4K
A Calculation of probabilities assuming correlation between events
Replies
9
Views
1K
A Calculation of average damage using analytical methods
Replies
3
Views
2K
A Joint Used to Show lack of Correlation?
Replies
43
Views
5K
I Do These Probability Conditions Imply Independence of Events A, B, and C?
Replies
7
Views
2K
I Find P(X+Y>1/2) for given joint density function
Replies
4
Views
3K
Conditional expectations related to count of event occurring kth time
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top