Problem with a proof in probability

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SUMMARY

The discussion centers on proving the inequality P(A and B|A) ≥ P(A and B|A or B) for two events A and B. The participant attempted to simplify the left side using the equation P(A and B|A) = P(A) . P(B|A), leading to the conclusion that P(A and B|A) equals P(A and B). However, confusion arose regarding the simplification of the right side, particularly in understanding the priority of conditions in probability. The solution involves applying the definition of conditional probability and comparing the numerators and denominators of both sides.

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  • Knowledge of basic probability equations, specifically P(A and B) and P(A|B)
  • Ability to manipulate algebraic expressions in the context of probability
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  • Study the properties of conditional probability, focusing on P(A|B) and P(B|A)
  • Learn about the law of total probability and its applications
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  • Practice solving inequalities involving conditional probabilities
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Students studying probability theory, mathematicians, and anyone involved in statistical analysis who seeks to deepen their understanding of conditional probabilities and their relationships.

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Homework Statement


I have to prove that for any two events A and B
P(A and B|A) [tex]\geq[/tex] P(A and B| A or B)

Homework Equations


P(A and B) = P(A) . P(B|A)
P(A|B) = (P(A) . P(B|A))/P(B)


The Attempt at a Solution


I tried to simplify the left side with this reasoning
P(A and B|A) = P(A) . P((B|A)|A)
= P(A) . P(B|A)
= P(A and B)

My reasoning for going from step 1 to two is that condition A is already fulfilled, and asking for it a second time is needless. My friend however, disagrees with this.
I am having problems with simplifying the right side because I don't know if there is a system of priorities in probability mathematics (e.g. "condition" has a priority over "and", "or" over "condition", ...
Any help in the right direction would be greatly welcomed.
 
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Remember that for any two events [tex]\mathcal{C}, \mathcal{D}[/tex] you have

[tex] \Pr(\mathcal{C} | \mathcal{D}) = \frac{\Pr(\mathcal{C} \cap \mathcal{D})}{\Pr{\mathcal{D}}}[/tex]

Use this idea to write out both of the probabilities you need to compare. If you do it correctly you should notice something about the numerators and the denominators, and those items will allow you to argue for the conclusion you need.
 

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