Max Value of $P(A \cap B): \min(P(A),P(B))

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SUMMARY

The maximum value of the intersection of two events, $P(A \cap B)$, is determined by the minimum of their individual probabilities, specifically $\min(P(A), P(B))$, when the sum of their probabilities exceeds one, i.e., $P(A) + P(B) > 1$. The relationship can be established using the equation $P(A \cap B) = P(A) + P(B) - P(A \cup B)$. By rearranging this equation, it is evident that $P(A \cap B)$ cannot exceed either $P(A)$ or $P(B)$, confirming the conclusion.

PREREQUISITES
  • Understanding of probability theory concepts, specifically event probabilities.
  • Familiarity with set operations in probability, including union and intersection.
  • Knowledge of the properties of probabilities, particularly the addition rule.
  • Ability to manipulate algebraic expressions involving probabilities.
NEXT STEPS
  • Study the addition rule of probabilities in detail.
  • Learn about the properties of independent and dependent events in probability.
  • Explore the concept of conditional probability and its applications.
  • Investigate the implications of the inclusion-exclusion principle in probability theory.
USEFUL FOR

This discussion is beneficial for students of probability theory, mathematicians, and anyone involved in statistical analysis or data science who seeks to deepen their understanding of event relationships and probability calculations.

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Suppose $A$ and $B$ are events with $P(A)+P(B)>1$. Show that the largest possible value of $P(A \cap B)$ is $ \min(P(A), P(B))$.

I suspect I'm supposed to use $P(A \cap B) = P(A)+P(B) -P(A \cup B)$ but I've no idea how.
 
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Rearrange your equation to:
$$P(A)-P(A\cap B)=P(A\cup B)-P(B)$$
I hope you now see that
$$P(A\cap B)\leq P(A)$$
Now finish.
 
johng said:
Rearrange your equation to:
$$P(A)-P(A\cap B)=P(A\cup B)-P(B)$$
I hope you now see that
$$P(A\cap B)\leq P(A)$$
Now finish.
Thank you. I get it now. :D
 
Last edited:

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