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anonymous

  • 4 years ago

P(EF') = P(E) - P(EF) Is it true or false? if its false how?

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  1. anonymous
    • 4 years ago
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    is this probability?

  2. anonymous
    • 4 years ago
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    yes

  3. anonymous
    • 4 years ago
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    \[P(E\cap F)=P(E)-F(E\cap F^c)\]?

  4. anonymous
    • 4 years ago
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    yeah so is it true?

  5. anonymous
    • 4 years ago
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    yeah it is true

  6. anonymous
    • 4 years ago
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    no not F(EF') its P(EF')

  7. anonymous
    • 4 years ago
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    first of all by previous exercise we know that \[E=(E\cap F)\cup (E\cap F^c)\]

  8. anonymous
    • 4 years ago
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    yeah that was as typo

  9. anonymous
    • 4 years ago
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    oh ok lol

  10. anonymous
    • 4 years ago
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    and since \[E\cap F\] and \[E\cap F^c\] are disjoint, the probability of their union is the sum of their probabilities, that is \[P((E\cap F)\cup (E\cap F^c))=P(E\cap F) +P (E\cap F^c)\]

  11. anonymous
    • 4 years ago
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    therefore since the sets are the same, you have \[P(E)=P(E\cap F) +P (E\cap F^c)\]

  12. anonymous
    • 4 years ago
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    ohh thank you

  13. anonymous
    • 4 years ago
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    if you think about what this says in english it is obvious. you are interested in the probability of E so you know you are in the set E. now if you are in E either you are in F or you are not in F those are the logical possibilities. so \[E=(E\cap F)\cup (E\cap F^c)\]

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