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Math Help - probability distributions - answer check

  1. #1
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    probability distributions - answer check

    A pair of dice is rolled. What are the odds in favor of each of the following events occuring?

    a) a sum of 7 turning up

    b) a sum of 11 turning up

    c) a sum of 7 or a sum of 11 turning up


    here is what i have so far..

    a) n(S) = 6x6=36

    E(rolling a 7) = {(1,6),(2,5),(4,3),(6,1),(5,2),(3,4)}
    n(E)= 6

    P(E)= 6/36 simplified 1/6

    so getting the odds would be..

    1/6 / 5/6 = 1/5

    therefore the odds are 1 to 5

    b) n(S) = 6x6=36

    E(rolling a 11) = {(6,5),(5,6)}
    n(E)= 2

    P(E)= 2/36 simplified 1/18

    so getting the odds would be..

    1/18 / 17/18 = 1/17

    therefore the odds are 1 to 17

    c) a sum of 7 or a sum of 11 turning up

    probability of rolling a 7 = 1/6

    probability of rolling a 11 = 1/18

    1/6x1/18 = 1/108

    P(E)/P(E') = 1/108/107/108 = 1/107

    therefore the odds are 1 to 107

    any help would be greatly appreciated.
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  2. #2
    MHF Contributor
    Joined
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    West Malaysia
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    Quote Originally Posted by ovi8 View Post
    A pair of dice is rolled. What are the odds in favor of each of the following events occuring?

    a) a sum of 7 turning up

    b) a sum of 11 turning up

    c) a sum of 7 or a sum of 11 turning up


    here is what i have so far..

    a) n(S) = 6x6=36

    E(rolling a 7) = {(1,6),(2,5),(4,3),(6,1),(5,2),(3,4)}
    n(E)= 6

    P(E)= 6/36 simplified 1/6

    so getting the odds would be..

    1/6 / 5/6 = 1/5

    therefore the odds are 1 to 5

    b) n(S) = 6x6=36

    E(rolling a 11) = {(6,5),(5,6)}
    n(E)= 2

    P(E)= 2/36 simplified 1/18

    so getting the odds would be..

    1/18 / 17/18 = 1/17

    therefore the odds are 1 to 17

    c) a sum of 7 or a sum of 11 turning up

    probability of rolling a 7 = 1/6

    probability of rolling a 11 = 1/18

    1/6x1/18 = 1/108

    P(E)/P(E') = 1/108/107/108 = 1/107

    therefore the odds are 1 to 107

    any help would be greatly appreciated.

    (a) and (b) look good . But for (c) , you have to add up the probabilities instead of multiplying them ie 1/6+1/18
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