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Math Help - (in)dependent events?

  1. #1
    Member courteous's Avatar
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    Question (in)dependent events?

    We have 4 cards on a table: \spadesuit K, \spadesuit Q, \heartsuit K and \heartsuit Q.
    We blindly choose one card.
    Define events as:
    A \equiv \spadesuit chosen
    B \equiv Q chosen
    C \equiv \heartsuit K \cup \spadesuit Q chosen

    Are following events independent?

    1. A \cap B
    2. A \cap C
    3. B \cap C
    4. A \cap B \cap C
    1. Yes.
    2. Yes: P(A \cap C)=P(A)*P(C|A)=\frac{1}{2}\frac{1}{2}==P(A)*P(C)=\  frac{1}{2}\frac{1}{2}
    3. Yes: P(B \cap C)=P(B)*P(C|B)=\frac{1}{2}\frac{1}{2}==P(B)*P(C)=\  frac{1}{2}\frac{1}{2} [Solutions say "No."]
    4. No: P(A \cap B \cap C)=P(A)*P(B|A)*P(C|A\cap B)=\frac{1}{2}\frac{1}{2}1\neq P(A)*P(B)*P(C)=\frac{1}{2}\frac{1}{2}\frac{1}{2}

    Please, also tell me whether points 2. and 4. are correct (for the right reasons)?

    Thank you!
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  2. #2
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    Hi Courteous,

    I think you are correct on all 4 parts.

    However, I think you have sometimes done things the hard way. Rather than computing P(A \cap C) as P(A) * P(C | A), for example, it's easier to observe that A \cap C is the event of drawing the queen of spades, 1 card out of 4, hence its probability is 1/4.
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  3. #3
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    I think some of the confusion in the OP comes from notation.
    Express the events in set notation: A = \left\{ {\spadesuit K,\spadesuit Q} \right\},\,B = \left\{ {\spadesuit Q,\heartsuit Q} \right\}\;\& \,C = \left\{ {\heartsuit K,\spadesuit Q} \right\}
    It becomes clear that P(A)=P(B)=P(C)=0.5.

    Now the question as to if events A~\&~C are independent comes down to asking if P\left( {A \cap C} \right)\mathop  = \limits^? P(A)P(C).
    But A\cap C=\left\{ {\spadesuit Q} \right\} which has probability 0.25.

    The question about A,~B,~\&~C being independent is actually vague.
    It is clear that  P\left( {A \cap B \cap C} \right) \ne P(A)P(B)P(C). Is that what is meant?

    I also do not understand the bit in red.
    If the solution set says that, it is wrong.
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  4. #4
    Member courteous's Avatar
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    Quote Originally Posted by Plato View Post
    The question about A,~B,~\&~C being independent is actually vague.
    It is clear that  P\left( {A \cap B \cap C} \right) \ne P(A)P(B)P(C). Is that what is meant?
    Instructions in OP are as given, so I know even less.

    I would like to make another "are (in)dependent?" problem:
    We have 8 cards: A, K, Q, J, 10, 9, 8, 7.
    We pick one.
    Events:
    A=\left\{A,K,Q,J\right\}
    B=\left\{A,9,8,7\right\}
    C=\left\{A,K,Q,10\right\}
    Are events A, B and C independent?
    P(A \cap B \cap C)=\frac{1}{8}==P(A)*P(B)*P(C)=\frac{1}{2}\frac{1}  {2}\frac{1}{2}
    Solutions say "No, they're not independent (even though P(A \cap B \cap C)=P(A) P(B) P(C))."
    Seems reasonable to me also (looking at events' definitions), but how to argue that, indeed they're not independent?

    Does independency require that all pairs are also independent (that, say, P(C|A)=P(C), which is obviously not true)?
    Last edited by courteous; August 25th 2010 at 11:56 PM.
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  5. #5
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    Quote Originally Posted by courteous View Post
    P(A \cap B \cap )=\frac{1}{8}==P(A)*P(B)*P(C)=\frac{1}{2}\frac{1}{  2}\frac{1}{2}
    Solutions say "No, they're not independent (even though P(A \cap B \cap C)=P(A) P(B) P(C))."
    Does independency require that all pairs are also independent (that, say, P(C|A)=P(C), which is obviously not true)?
    Yes, you have answered your own question.
    Each subcollection has to be independent. So each pair must be independent.
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