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Math Help - dice with dependence events

  1. #1
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    dice with dependence events

    A fair six sided die is tossed 10 times and the resulting sequence of outcomes is recorded.

    Let E be the event that the sequence of 10 outcomes contains exactly 4 ones and let G be the event that the sequence of 10 outcomes contains exactly 6 threes. Find P(E U G)
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    Quote Originally Posted by digitalis77 View Post
    A fair six sided die is tossed 10 times and the resulting sequence of outcomes is recorded.

    Let E be the event that the sequence of 10 outcomes contains exactly 4 ones and let G be the event that the sequence of 10 outcomes contains exactly 6 threes. Find P(E U G)
    P(E) = {10 \choose 4} \left( \frac{1}{6} \right) ^4 \left( \frac{5}{6} \right) ^6

    P(G) = {10 \choose 6} \left( \frac{1}{6} \right) ^6 \left( \frac{5}{6} \right) ^4
    Last edited by janvdl; May 9th 2008 at 12:30 PM.
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  3. #3
    Moo
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    Quote Originally Posted by janvdl View Post
    P(E \cup G) = P(E) + P(G)
    Hi !

    Isn't that P(E \cup G) = P(E) + P(G) - P(E \cap G) ?

    E and G can happen together : 4x1 and 6x3
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    Quote Originally Posted by janvdl View Post
    But they are independent. Rolling a 1 is not going to influence the rolling of a 3.
    I think Moo is correct on this one. In each one of the probabilities of rolling four ones and six threes are included the possibilities of rolling four ones and six threes, which total \frac{10!}{4!6!}, and are being counted twice.
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    Moo
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    Quote Originally Posted by janvdl View Post
    But they are independent. Rolling a 1 is not going to influence the rolling of a 3.
    I thought the independence could be used in P(E \cap G)=P(E)\cdot P(G) if E and G are independent.

    Are you sure they are ? If you get 4 ones in a first time, you won't have as many chances as before to get 6 threes.. will you ?
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    Bar0n janvdl's Avatar
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    Okay, seeing as both of you are convinvced of your answers, I will delete the last part of mine. Thanks for the spot.
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    Quote Originally Posted by Moo View Post
    I thought the independence could be used in P(E \cap G)=P(E)\cdot P(G) if E and G are independent.
    Moo, they are not independent!
    P(E \cap G) = \binom{10}{ 4} \left( {\frac{1}{{6^{10} }}} \right).
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    Quote Originally Posted by Plato View Post
    Moo, they are not independent!
    P(E \cap G) = \binom{10}{ 4} \left( {\frac{1}{{6^{10} }}} \right).
    Oh well, all 3 of us were wrong. Thanks Plato.
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  9. #9
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    Quote Originally Posted by janvdl View Post
    Oh well, all 3 of us were wrong. Thanks Plato.
    Moo's first post was correct. I was trying to say what Plato said, that you have to take into account the intersection of the event probabilities.
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  10. #10
    Moo
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    Quote Originally Posted by Plato View Post
    Moo, they are not independent!
    P(E \cap G) = \binom{10}{ 4} \left( {\frac{1}{{6^{10} }}} \right).
    I didn't say they were oO
    I said that independence was used in this sort of formula


    Are you sure they are ? If you get 4 ones in a first time, you won't have as many chances as before to get 6 threes.. will you ?
    This was about independence

    Nevermind...one day, I'll learn how to speak clearly lol
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