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Independence (Probability)

  1. Oct 23, 2013 #1
    The question in my book is posed as:

    Find an example in which [itex]P(AB) < P(A)P(B)[/itex].
    The answer is given as a coin toss letting A be the event of obtaining heads and B being the even of obtaining tails. I'm confused as to how that's working.

    My understanding is no doubt 'off', but I thought it was
    [tex]
    P(\frac{1}{2} * \frac{1}{2}) = \frac{1}{2} * \frac{1}{2}
    [/tex]
    because the probabilities for both are [itex]\frac{1}{2}[/itex]. So they are equivalent at [itex]\frac{1}{4}[/itex]. Clearly I'm incorrect, can someone point out why?

    Thanks.
     
  2. jcsd
  3. Oct 23, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    AB is shorthand for the cases where A is true AND B is true. Wouldn't it be true that the probability of getting heads AND tails is 0?
     
  4. Oct 23, 2013 #3
    That clarifies everything. Thank you.
     
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