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How to Determine wether A and B are Mutually Exclusive and Independent Watch

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    I've been given the information:
    P(A) = 0.25 P(B) = 0.4 and P(A'nB') = 0.45

    With this information how do I determine if A and B are Mutually Exclusive and/or Independent?
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    (Original post by AdamChris)
    I've been given the information:
    P(A) = 0.25 P(B) = 0.4 and P(A'nB' = 0.45

    With this information how do I determine if A and B are Mutually Exclusive and/or Independent?
    Use the Event Independence Test.

    Two events are independent if, and only if, P(A\lvert B)=P(A), and P(A\cap B)=P(A)P(B) also follows from it.

    \displaystyle P(A\lvert B) = \frac{P(A\cap B)}{P(B)}
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    (Original post by RDKGames)
    Use the Event Independence Test.

    Two events are independent if, and only if, P(A\lvert B)=P(A), and P(A\cap B)=P(A)P(B) also follows from it.

    \displaystyle P(A\lvert B) = \frac{P(A\cap B)}{P(B)}
    Is P(AnB) the same as 1- P(A'nB')?
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    (Original post by AdamChris)
    Is P(AnB) the same as 1- P(A'nB'?
    Yes. I think. Not the best at probability so I'm unsure.

    Also if two events are mutually exclusive then P(A \cap B) = 0 (ie: the probability of A and B happening at the same time is 0)
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    (Original post by AdamChris)
    Is P(AnB) the same as 1- P(A'nB'?
    No.  A'\cap B' = (A\cup B)' .
    You need to find  P(A\cap B ) to see the events are independent.
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    As above, note that P(A \cup B) = P(A) + P(B) - P(A \cap B) So P(A \cup B) = 0.65 -  P(A \cap B)

    and P(A' \cap B') = 1 - P(A \cup B) \Rightarrow can you take it from there?
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    In addition to what others have stated above, note De Morgan's Law:

     A' \cap B' = (A \cup B)'.

    (Source: http://www.math-only-math.com/proof-...rgans-law.html )

    Find  P(A \cap B) and use the formula for independence of two events and what it means if two events are mutally exclusive to each other.
 
 
 
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