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Math Help - Two fair dice are thrown. Prob of event A.

  1. #1
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    Two fair dice are thrown. Prob of event A.

    Hi to all.

    Simple question but answer seems to be coming wrong.

    Question says:

    Two fair dice are thrown.

    (i) Event A is ‘the scores differ by 3 or more’. Find the probability of event A.

    (ii) Event B is ‘the product of the scores is greater than 8’. Find the probability of event B.

    When it says the scores differ by 3 or more, what I understand is that the difference of the results should be 3 or more. This means 3,4,5 and 6. So, won't we have the following pairs:

    For difference of 3: (6,3) (5,2) (4,1) and Reverse.
    For difference of 4: (6,2) (5,1) and Reverse.
    For difference of 5: (6,1) and Reverse.
    For difference of 6: None.

    So we'd have (1/6)^6 + (1/6)^4 + (1/6)^2 ...? The answer I get is 1333/46656 which appears to be wrong. Maybe I didn't get the question right. So any help here?

    Part (ii) likely has the same problem.
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  2. #2
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    Quote Originally Posted by SolCon View Post
    Two fair dice are thrown.
    (i) Event A is ‘the scores differ by 3 or more’. Find the probability of event A.
    (ii) Event B is ‘the product of the scores is greater than 8’. Find the probability of event B.
    For difference of 3: (6,3) (5,2) (4,1) and Reverse.
    For difference of 4: (6,2) (5,1) and Reverse.
    For difference of 5: (6,1) and Reverse.
    For difference of 6: None.
    That gives us only 12 possible successful outcomes,
    So the probability is \frac{12}{36}.
    Now you show us part (ii).
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  3. #3
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    Thanks for the reply.

    I didn't realize that we were not supposed to use each individual result's probability (all having the same, i.e 1/6). So whenever it says "probability of Event This or that", I'm to assume it only means the possible outcomes under condition given from all possible outcomes? In this case, the outcomes possible under condition given is 12, and the total possible outcomes being 36?

    As for part(ii), as I've now understood what the question is asking, I've arranged the numbers as follows where the condition is that multiplying the results should give us a score of more than 8:

    First dice throw / Second dice throw(s)
    2/ 5,6
    3/ 3,4,5,6
    4/ 3,4,5,6
    5/ 2,3,4,5,6
    6/ 2,3,4,5,6

    Which, when counted, appears to be 20 results, so the answer should be 20/36?
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  4. #4
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    Correct.
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  5. #5
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    haha!.

    Well, thanks for the help.
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  6. #6
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    Hello, SolCon!

    Don't strain-your-brain!
    List all the outcomes and count the ones you want.


    \text{Two fair dice are thrown.}

    There are thirty-six possible outcomes:

    . . \begin{array}{cccccc} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\<br />
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\<br />
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\<br />
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\<br />
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\<br />
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}




    (1) Event \,A is "the scores differ by 3 or more".
    . . . Find the probability of event \,A.

    There are twelve outcomes for event \,A.

    \begin{array}{cccccc} (1,1) & (1,2) & (1,3) & {\bf(1,4)} & {\bf(1,5)} & {\bf(1,6)} \\<br />
(2,1) & (2,2) & (2,3) & (2,4) & {\bf(2,5)} & {\bf(2,6)} \\<br />
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & {\bf(3,6)} \\<br />
{\bf(4,1)} & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\<br />
{\bf(5,1)} & {\bf(5,2)} & (5,3) & (5,4) & (5,5) & (5,6) \\<br />
{\bf(6,1)} & {\bf(6,2)} & {\bf(6,3)} & (6,4) & (6,5) & (6,6) \end{array}


    \text{Therefore: }\:P(A) \:=\:\dfrac{12}{36} \:=\:\dfrac{1}{3}




    (2) Event \,B is "the product of the scores is greater than 8".
    . . . Find the probability of event \,B.

    There are twenty outcomes for event \,B.

    \begin{array}{cccccc} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\<br />
(2,1) & (2,2) & (2,3) & (2,4) & {\bf(2,5)} & {\bf(2,6)} \\<br />
(3,1) & (3,2) & {\bf(3,3)} & {\bf(3,4)} & {\bf(3,5)} & {\bf(3,6)} \\<br />
(4,1) & (4,2) & {\bf(4,3)} & {\bf(4,4)} & {\bf(4,5)} & {\bf(4,6)} \\<br />
(5,1) & {\bf(5,2)} & {\bf(5,3)} & {\bf(5,4)} & {\bf(5,5)} & {\bf(5,6)} \\<br />
(6,1) & {\bf(6,2)} & {\bf(6,3)} & {\bf(6,4)} & {\bf(6,5)} & {\bf(6,6)} \end{array}


    \displaystyle\text{Therefore: }\:P(B) \;=\;\frac{20}{36} \;=\;\frac{5}{9}

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