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Math Help - Are these events independent?

  1. #1
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    Are these events independent?

    Choose a number X from {1,2,3,4,5}. Now choose a number from the subset {1.....X}. Call this second number Y.

    P{X|Y=1} = 1/5
    P{X|Y=2| = 1/4
    P{X|Y=3} = 1/3
    P{X|Y=4} = 1/2
    P{X|Y=5} = 1

    Are these probabilities correct? These event are not independent?
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  2. #2
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    Quote Originally Posted by lord12 View Post
    Choose a number X from {1,2,3,4,5}. Now choose a number from the subset {1.....X}. Call this second number Y.

    P{X|Y=1} = 1/5
    P{X|Y=2| = 1/4
    P{X|Y=3} = 1/3
    P{X|Y=4} = 1/2
    P{X|Y=5} = 1

    Are these probabilities correct? These event are not independent?
    No its P(X|Y=1) = 1

    P(X|Y=2) = 4/5

    P(X|Y=3) = 3/5

    etc...

    still not independent because  P(X|Y) \neq P(X) .
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  3. #3
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    Quote Originally Posted by particlejohn View Post
    No its P(X|Y=1) = 1

    P(X|Y=2) = 4/5

    P(X|Y=3) = 3/5

    etc...

    still not independent because  P(X|Y) \neq P(X) .
    \Pr(X = 1 | Y = 1) \neq \Pr(X = 1) so X and Y are clearly dependent.

    I'm not sure what Pr(X | Y = 1) = 1 means ......

    \Pr(X = 1 | Y = 1) = \frac{ \Pr(Y = 1 | X = 1) \cdot \Pr(X = 1) }{\Pr(Y = 1)} = \frac{ \frac{1}{5}}{\Pr(Y = 1)}

    where

    \Pr(Y = 1) = \Pr(Y = 1 | X = 1) \cdot \Pr(X = 1) + \Pr(Y = 1 | X = 2) \cdot \Pr(X = 2) + \Pr(Y = 1 | X = 3) \cdot \Pr(X = 3)

    + \Pr(Y = 1 | X = 4) \cdot \Pr(X = 4) + \Pr(Y = 1 | X = 5) \cdot \Pr(X = 5)


     = (1) \, \left(\frac{1}{5}\right) + \left(\frac{1}{2}\right) \cdot \left(\frac{1}{5}\right) + \left(\frac{1}{3}\right) \cdot \left(\frac{1}{5}\right) + \left(\frac{1}{4}\right) \cdot \left(\frac{1}{5}\right) \left(\frac{1}{5}\right) \cdot \left(\frac{1}{5}\right)


    = \frac{1}{5} + \frac{1}{10} + \frac{1}{15} + \frac{1}{20} + \frac{1}{25} = \frac{137}{300}.


    Therefore
    \Pr(X = 1 | Y = 1) = \frac{ \frac{1}{5}}{\frac{137}{300}} = \frac{60}{137}.


    Obviously \Pr(X = 2 | Y = 1) = \frac{ \frac{1}{10}}{\frac{137}{300}} = \frac{30}{137}.

    And so on .....

    Does Pr(X | Y = 1) mean Pr(X = 1, 2, 3, 4 or 5 | Y = 1) .....?
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