Is P(A,B|C) = P(A|C) P(B|C), if P(A,B) = P(A)P(B)?

Click For Summary
SUMMARY

The assertion that P(A,B|C) = P(A|C) P(B|C) when P(A,B) = P(A)P(B) is false. This was demonstrated using the example of two independent fair coins, where A represents the first coin landing heads, B the second coin landing heads, and C the event that at least one coin shows heads. The calculations show that while P(A) = P(B) = 1/2 and P(A,B) = 1/4, the conditional probabilities yield P(A|C) = P(B|C) = 2/3, leading to P(A,B|C) = 1/3, which does not equal P(A|C)P(B|C) = 4/9. This indicates a dependence created by the condition C.

PREREQUISITES
  • Understanding of conditional probability
  • Familiarity with Bayes' theorem
  • Knowledge of independent events in probability
  • Basic concepts of probability distributions
NEXT STEPS
  • Study the implications of conditional independence in probability theory
  • Learn about Bayes' theorem applications in real-world scenarios
  • Explore examples of dependent and independent events in probability
  • Investigate the concept of joint probability distributions
USEFUL FOR

Mathematicians, statisticians, data scientists, and anyone involved in probability theory or statistical analysis will benefit from this discussion.

natski
Messages
262
Reaction score
2
As stated in my subject line, I know that P(A|B) = P(A) and P(B|A) = P(B), i.e. A and B are separable as P(A,B) = P(A) P(B). I strongly suspect that this holds with a conditional added, but I can't find a way to formally prove it... can anyone prove this in a couple of lines via Bayes' rules? This is not a homework question, but part of my research and I can't find the answer anywhere.

Thanks to anyone who can help in advanced!
natski
 
Physics news on Phys.org
No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but P(A|C) = P(B|C) = 2/3 and P(A,B|C) = 1/3 \neq P(A|C)P(B|C) = 4/9
 
  • Like
Likes   Reactions: FactChecker
Citan Uzuki said:
No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but P(A|C) = P(B|C) = 2/3 and P(A,B|C) = 1/3 \neq P(A|C)P(B|C) = 4/9
Even more obvious is C= exactly one coin is a head. Then the condition C forces a complete dependence between A and B.
 

Similar threads

High School Why did it suddenly become subtractive? (Example of Bayes’ Theorem)
  • · Replies 9 ·
Replies
9
Views
2K
High School How to calculate this conditional probability?
  • · Replies 7 ·
Replies
7
Views
804
Undergrad Do These Probability Conditions Imply Independence of Events A, B, and C?
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad Ways to put n balls in m boxes such that exactly k boxes are empty?
  • · Replies 29 ·
Replies
29
Views
4K
Undergrad How Can Bayes' Formula Help Understand Price Increases with Overseas Expansion?
  • · Replies 5 ·
Replies
5
Views
2K
High School Are proofs needed for definitions? Conditional probabilities
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Optimizing Defective Rates with Bayes' Formula
  • · Replies 17 ·
Replies
17
Views
2K
MHB Rules of Inference for Proving p→(p→q)→(p→q)
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad ##(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}## and its isomorphism?
  • · Replies 31 ·
2
Replies
31
Views
2K
Undergrad Understanding the Difference Between P (A, B) and P (B, A)
  • · Replies 3 ·
Replies
3
Views
3K