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Math Help - 7th grader needs help

  1. #1
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    7th grader needs help

    ok im doing probability and i was absent so i have no notes and my teacher explains stuff way bad.

    the question is

    ''a die is rolled whats the probability?''

    p(odd)

    and on another problem it asks me to draw a chart for when two dice are rolled and probabilitys.

    what kind of chart?

    the questions are

    P(both odd)
    P(both prime)
    P(sum of 7)
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  2. #2
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    =s

    help plz
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  3. #3
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    Hello,Nostradamus!

    A die is rolled. What's the probability that the number is odd?

    With one die, there are six outcomes: . 1, 2, 3, 4, 5, 6

    Three of them \{1,3,5\} are odd.

    Therefore: . P(\text{odd}) \:=\:\frac{3}{6}\:=\:\frac{1}{2}



    Draw a chart for when two dice are rolled.
    . . \begin{array}{cccccc}(1,1)&(1,2)&(1,3)&(1,4)&(1,5)  &(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)  &(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\(4,1)&(4,2)&(4,3)&  (4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(  5,6) \\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}


    Find these probabilities:
    P(both odd)
    P(both prime)
    P(sum of 7)
    There are 36 possible outcomes.


    Both odd
    There are nine cases: . \begin{array}{ccc}(1,1)&(1,3)&(1,5)\\(3,1)&(3,3)&(  3,5)\\(5,1)&(5,3)&(5,5)\end{array}

    P(\text{both odd}) \:=\:\frac{9}{36} \:=\:\frac{1}{4}


    Both prime
    There are nine cases: . \begin{array}{ccc}(2,2)&(2,3)&(2,5)\\(3,2)&(3,3)&(  3,5)\\(5,2)&(5,3)&(5,5)\end{array}

    P(\text{both prime}) \:=\:\frac{9}{36} \:=\:\frac{1}{4}


    Sum of 7

    There are six cases: . \begin{array}{cccccc}(1,6)&(2,5)&(3,4)&(4,3)&(5,2)  &(6,1)\end{array}

    P(\text{sum of 7}) \:=\:\frac{6}{36}\:=\:\frac{1}{6}

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  4. #4
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    p(odd)

    there is 6 possibilitys 3 are odd (1-3-5), 3 are even (2-4-6)

    so it is 50% or 1/2
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