Conditional Probabilities Complementary Proof

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SUMMARY

The discussion centers on the proof of conditional probabilities, specifically the relationship between P(A|B) and P(~A|B). The user seeks clarification on how to simplify the expression P(A&B) + P(~A&B). It is established that P(A|B) can be expressed as P(A&B)/P(B) and P(~A|B) as P(~A&B)/P(B), leading to the conclusion that P(A&B) + P(~A&B) equals P(B). This highlights the foundational principles of conditional probability and its complementary nature.

PREREQUISITES
  • Understanding of conditional probability
  • Familiarity with probability notation (e.g., P(A), P(B))
  • Basic knowledge of set theory and events
  • Ability to manipulate algebraic expressions in probability
NEXT STEPS
  • Study the derivation of Bayes' Theorem
  • Learn about joint and marginal probabilities
  • Explore the law of total probability
  • Investigate applications of conditional probabilities in real-world scenarios
USEFUL FOR

Students of statistics, data scientists, and anyone interested in mastering the concepts of probability theory and its applications in various fields.

rbzima
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I'm having trouble seeing how this works out. It's blatantly obvious that this is true, but somehow I can't seem to get anywhere on paper with it to simplify it down to anything. Any help would be greatly appreciated!

[tex]P\left(A\right|B)=1-P\left(not A\right|B)[/tex]
 
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P(A|B) = P(A&B)/P(B)
P(~A|B) = P(~A&B)/P(B)

What is P(A&B) + P(~A&B) = ?
 

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