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adikcomel

  • one year ago

help me please.. Show that if P (A) >0 , then P(A ∩ B | A) ≥ P(A∩B| A ∪B )

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  1. Zarkon
    • one year ago
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    use the definition of conditional probability and the fact that \[P(A)\le P(A\cup B)\]

  2. adikcomel
    • one year ago
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    okay i'll try first :)

  3. adikcomel
    • one year ago
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    P( A intersect B | A) = P( (A intersect B) intersect A)/P(A) = P(A intersect B)/P(A) P( A intersect B | A union B) = P( {A intersect B} intersect {A union B})/P(A union B) ..= P(A intersect B)/P(A union B). As the numerators for the two expressions are equal, while the denominator defining P( A intersect B | A union B) is larger than the denominator defining P( A intersect B | A ), it follows that P( A intersect B | A) >= P( A intersect B | A union B)

  4. adikcomel
    • one year ago
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    is that true ?

  5. Zarkon
    • one year ago
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    yes

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