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Math Help - 2 Questions about probability+De Morgan laws

  1. #1
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    2 Questions about probability+De Morgan laws

    about the first Question is known that A and B are Foreign events.
    and i asked to find P(A).


    about the second Question is known that A and B are event and i asked to decide what value y cannot get.



    i tried with the De Morgan laws but i am not sure if i correct in my way.
    Attached Thumbnails Attached Thumbnails 2 Questions about probability+De Morgan laws-25985616560438177439.jpg  
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  2. #2
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    What in the world are Foreign events?
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  3. #3
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    Quote Originally Posted by ZOOZ View Post
    about the first Question is known that A and B are Foreign events.
    and i asked to find P(A).


    about the second Question is known that A and B are event and i asked to decide what value y cannot get.



    i tried with the De Morgan laws but i am not sure if i correct in my way.
    Until we know what you mean by 'foreign' events, there is not enough information to answer these questions.
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  4. #4
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    oh well yes i knew is will be wrong to say Foreign events, so what i mean is A and B are mutually exclusive events.
    sorry about the english.


    thanks.
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  5. #5
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    Quote Originally Posted by ZOOZ View Post
    oh well yes i knew is will be wrong to say Foreign events, so what i mean is A and B are mutually exclusive events.
    sorry about the english.
    Well then that means A^c\cap B=B~\&~A\cap B^c=A.
    Use the given.

    Moreover, P(A^c\cap B^c)=1-P(A\cup B).
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  6. #6
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    thank you for your answer, but i still not sure how can i know what P(A) can be.
    i just know that P(A ∩ B) = 0 but i am not sure if is help me here.

    can you please give me more tip to solve this question?

    thanks again.
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  7. #7
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    Quote Originally Posted by ZOOZ View Post
    thank you for your answer, but i still not sure how can i know what P(A) can be.
    i just know that P(A ∩ B) = 0 but i am not sure if is help me here.
    can you please give me more tip to solve this question?
    From the given: 2P(A)=2P(A\cap B^c)=2P(B\cap A^c)=2P(B).

    And P(A^c\cap B^c)=1-P(A)-P(B)

    That is all I will do.
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  8. #8
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    Quote Originally Posted by ZOOZ View Post
    thank you for your answer, but i still not sure how can i know what P(A) can be.
    i just know that P(A ∩ B) = 0 but i am not sure if is help me here.

    can you please give me more tip to solve this question?

    thanks again.
    Draw a Karnaugh Table:

    \begin{tabular}{l | c | c | c} & A & A' & \\ \hline B & 0 & x & x\\ B' & x & 2x & 3x \\ \hline & x & 3x & 1 \\ \end{tabular}

    It is simple to solve for x and hence get Pr(A).

    Quote Originally Posted by ZOOZ View Post
    [snip]
    about the second Question is known that A and B are event and i asked to decide what value y cannot get.

    i tried with the De Morgan laws but i am not sure if i correct in my way.
    As I said earlier, unless you say what sort of events A and B are, this question cannot be answered. But I suggest you draw a Karnaugh Table and make an attempt at it before posting your clarification. Then you can say what you've done and where you get stuck.
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  9. #9
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    ok for the 1 question i got P(A)=1/4.
    0.5x+1.5x=1

    x=0.5

    P(A)=x=0.5*0.5=0.25

    P(B)=0.5x=0.5*0.5=0.25
    do i correct?
    thanks to both of you.
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  10. #10
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    Quote Originally Posted by ZOOZ View Post
    ok for the 1 question i got P(A)=1/4.
    0.5x+1.5x=1

    x=0.5

    P(A)=x=0.5*0.5=0.25

    P(B)=0.5x=0.5*0.5=0.25
    do i correct?
    thanks to both of you.
    Yes, Pr(A) = Pr(B) = 1/4.

    You still haven't said what sort of events A and B are in question 2.
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  11. #11
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    well about question 2 it given that A and B are just events.
    is not say's if it an mutually exclusive events.
    so i think is just events in Sample space.

    thanks.
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  12. #12
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    Quote Originally Posted by ZOOZ View Post
    well about question 2 it given that A and B are just events.
    is not say's if it an mutually exclusive events.
    so i think is just events in Sample space.

    thanks.
    In that case there is not enough information to get a numerical answer.
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