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mathmath333

  • one year ago

Probablity question

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  1. mathmath333
    • one year ago
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    Two events A and B will be independent, if (A) A and B are mutually exclusive (B) (B) P(A′B′) = [1 – P(A)] [1 – P(B)] (C) P(A) = P(B) (D) P(A) + P(B) = 1

  2. misty1212
    • one year ago
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    i don't see the correct answer there

  3. misty1212
    • one year ago
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    independent means \[P(A|B)=P(A)\]

  4. mathmath333
    • one year ago
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    ?

  5. Zarkon
    • one year ago
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    the answer is there

  6. misty1212
    • one year ago
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    or \[P(A\cap B)=P(A)P(B)\] guess we need to look more carefully then huh?

  7. misty1212
    • one year ago
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    ooh is that B just \[P(A'\cap B')=(1-P(A))(1-P(B))\]?

  8. Zarkon
    • one year ago
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    yes

  9. misty1212
    • one year ago
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    sorry got confused with the proliferation of B's there

  10. misty1212
    • one year ago
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    @Zarkon can explain why this one is the correct one

  11. Zarkon
    • one year ago
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    well I assume that is what they meant

  12. misty1212
    • one year ago
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    process of elimination makes that what they meant

  13. freckles
    • one year ago
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    I think demorgan's theorem would be helpful

  14. freckles
    • one year ago
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    \[A' \text{ and } B'=(A \text{ or } B)'\]

  15. freckles
    • one year ago
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    \[P((A \text{ or } B)')=1-P(A \text{ or } B) \] then you can use that one thing for P(A or B)=P(A)+P(B)-P(A and B)

  16. freckles
    • one year ago
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    and misty said earlier P(A and B)=PA) P(B) since A and B are independent

  17. Zarkon
    • one year ago
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    P(A′B′) = [1 – P(A)] [1 – P(B)]=1-P(A)-P(B)+P(A)P(B) -1+P(A)+P(B)+P(A'B')=P(A)P(B) -1+(1-P(A'))+(1-P(B'))+P(A'B')=P(A)P(B) 1-(P(A')+P(B')-P(A'B')=P(A)P(B) 1-P(A'\(\cup\)B')=P(A)P(B) P(AB)=P(A)P(B)

  18. misty1212
    • one year ago
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    i think you only need to know that \[1-P(A)=P(A')\]

  19. anonymous
    • one year ago
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    like old home week today!

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