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Math Help - tetrahedron die probability

  1. #1
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    tetrahedron die probability

    A fair tetrahedron die has four faces , each with digit 1,2,3,4 . THe die is tossed three times. The sum of three visible surfaces is recorded as T at each toss . The events A , B and C are defined as follows .

    A= The score T at the first toss is 7 .
    B=SUm of first 2 scores is less than 15
    C=Product of the three scores is a multiple of 3 .

    FInd P(A) , P(B) and P(C)

    For P(A) , It can only (1,3,4) and this can be arranged in 3!=6 ways so
    P(A)=6/24=1/4

    For P(B) , its either (6,8) , (8,6) or (6,6)

    6=(1,2,3) , 8=(1,3,4)

    so p(B)=3 x (3!/4!)(3!/4!)=3/16 but the answer given is 3/8
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    A fair tetrahedron die has four faces , each with digit 1,2,3,4 . THe die is tossed three times. The sum of three visible surfaces is recorded as T at each toss . The events A , B and C are defined as follows .

    A= The score T at the first toss is 7 .
    B=SUm of first 2 scores is less than 15
    C=Product of the three scores is a multiple of 3 .

    FInd P(A) , P(B) and P(C)

    For P(A) , It can only (1,3,4) and this can be arranged in 3!=6 ways so
    P(A)=6/24=1/4

    For P(B) , its either (6,8) , (8,6) or (6,6)

    6=(1,2,3) , 8=(1,3,4)

    so p(B)=3 x (3!/4!)(3!/4!)=3/16 but the answer given is 3/8
    Don't overlook rolling two 7's (7=1+2+4).
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  3. #3
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    For A

    There is only 1 way to score a 7 on the first throw,
    which is 4, 2 and 1 if all are visible.

    As there are \binom{4}{3} ways to select 3 sides from the 4,

    then P(A)= \frac{1}{\binom{4}{3}}<br />
    For B

    the \binom{4}{3} possibilities are 4+3+2=9, 4+3+1=8, 4+2+1=7, 3+2+1=6

    Since 9+6=15, we cannot have 9.
    Since 8+7=15, only 8+6 and 6+8 contain 8, but 7 can pair with 6 or itself,
    giving 7+7, 7+6 and 6+7.
    The last one is 6+6.

    That's 6 options, so the probability is \frac{6}{\binom{4}{3}^2}

    For C

    the product of the 3 scores will be a multiple of 3 if 6 or 9 came up.
    6x and 9x are multiples of 3, for x natural.

    Hence, the product will not be a multiple of 3 if only 8 and 7 come up.
    888, 777 are two.
    All 3 arrangements of two 8's and a 7 and all 3 arrangements of two 7's and an 8 complete the list, so the probability is

    1-\frac{8}{4^3}
    Last edited by Archie Meade; January 17th 2010 at 05:30 AM. Reason: typo
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  4. #4
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    Hello, thereddevils!

    We don't need factorials for this problem.


    A fair tetrahedron die has four faces, each with digit 1,2,3,4.
    THe die is tossed three times.
    The sum of three visible surfaces is recorded as T at each toss.

    The events A , B and C are defined as follows .
    . . A: the score T at the first toss is 7
    . . B: the sum of first 2 scores is less than 15
    . . C: the product of the three scores is a multiple of 3.

    FInd: . P(A),\; P(B)\text{, and }P(C)
    On one roll, we have these outcomes:

    . . \begin{array}{ c c c }<br />
\text{Visible}& \text{Sum}& \\<br />
\text{faces} & (T) & \text{Prob.} \\ \hline \\[-3mm]<br />
1,2,3 & 6 & \dfrac{1}{4} \\ \\[-3mm]<br />
1,2,4 & 7 & \dfrac{1}{4} \\ \\[-3mm]<br />
1,3,4 & 8 & \dfrac{1}{4} \\ \\[-3mm]<br />
2,3,4 & 9 & \dfrac{1}{4} \\ \\[-3mm] \hline\end{array}


    (a)\;P(A) \;=\;P(\text{sum is 7}) \;=\;\frac{1}{4}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    With 2 rolls, there are: . 4^2 = 16 outcomes.


    If the sum is less than 15, we have:

    . . \begin{array}{ccc}<br />
\text{Rolls} & \text{Sum} & \text{Prob.} \\ \hline \\[-3mm]<br />
(6,6) & 12 & \dfrac{1}{16} \\ \\[-3mm]<br />
(6,7), (7,6) & 13 & \dfrac{2}{16} \\ \\[-3mm]<br />
(6,8),(7,7),(8,6) & 14 & \dfrac{3}{16} \\ \\[-3mm]\hline  \end{array}


    (b)\;P(B) \;=\; P(\text{sum is less than 15}) \;=\;\frac{1}{16} + \frac{2}{16} + \frac{3}{16} \;=\;\frac{6}{16} \;=\;\frac{3}{8}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    With three rolls, there are: . 4^3 = 64 outcomes.

    If the product of three rolls is a multiple of 3,
    . . at least one of the rolls is a multiple of 3.

    How many are not a multiple of 3?
    . . They must be comprised of 7's and/or 8's.

    . . \begin{array}{cc}<br />
\text{Rolls} & \text{Prob.} \\ \hline \\[-3mm]<br />
(7,7,7) & \dfrac{1}{64} \\ \\[-3mm]<br />
(7,7,8) & \dfrac{3}{64} \\ \\[-3mm]<br />
(7,8,8) & \dfrac{3}{64} \\ \\[-3mm]<br />
(8,8,8) & \dfrac{1}{64} \\ \\[-3mm] \hline \end{array}


    Hence: . P(not\text{ multiple of 3}) \;=\;\frac{1}{64} + \frac{3}{64} + \frac{3}{64} + \frac{1}{64} \;=\;\frac{8}{64} \;=\;\frac{1}{8}

    Therefore: . P(\text{multiple of 3}) \;=\;1-\frac{1}{8} \;=\;\frac{7}{8}

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