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Math Help - dice probability question

  1. #1
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    dice probability question

    A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is
    four times the probability of getting 2 even numbers. Find the probability that a single throw of the
    die results in an even number.


    Answer: 2/3

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  2. #2
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    Quote Originally Posted by ipokeyou View Post
    A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is
    four times the probability of getting 2 even numbers. Find the probability that a single throw of the
    die results in an even number.


    Answer: 2/3

    Thanks
    Hi ipokeyou,

    This is a binomial situation.
    We either get an even number or an odd number on any throw.
    We don't need to examine the specific 6 numbers themselves.

    There are \binom{8}{3} ways to get 3 even outcomes in 8 throws (doesn't matter which even number) and \binom{8}{2} ways to get 2 even outcomes.

    Therefore

    \binom{8}{3}p^3(1-p)^5=4\binom{8}{2}p^2(1-p)^6

    since 3 even throws must go with 5 odd throws etc
    and p=probability of an even result, 1-p=probability of an odd result.

    Hence,

    \frac{8!}{5!3!}p^3(1-p)^5=4\frac{8!}{6!2!}p^2(1-p)^6

    Diving both sides by p^2(1-p)^5

    \frac{8(7)6}{6}p=\frac{4(8)7}{2}(1-p)

    56p=112(1-p)

    56p=112-112p

    168p=112

    p=\frac{112}{168}=\frac{28}{42}=\frac{4}{6}=\frac{  2}{3}
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  3. #3
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    thanks for your help and quick response!!
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  4. #4
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    Hello, ipokeyou!

    A slight variation . . .


    A die is loaded in such a way that in 8 throws of the die, the probability
    of getting 3 even numbers is four times the probability of getting 2 even numbers.
    Find the probability that a single throw of the die results in an even number.

    Answer: 2/3

    Let: . \begin{array}{ccc}p &=& P(\text{even}) \\ q &=& P(\text{odd}) \end{array}\quad\text{ where }p+q\,=\,1


    We have: . \begin{array}{ccccc}P(\text{3 even)} &=& {8\choose3}p^3q^5 &=& 56p^3q^5\\ \\[-3mm]P(\text{2 even}) &=& {8\choose2}p^2q^6 &=& 28p^2q^6\end{array}


    . . Then: . 56p^3q^5 \:=\:4\cdot28p^2q^6 \quad\Rightarrow\quad p \:=\:2q


    Since p+q\:=\:1, we have: . p = \tfrac{2}{3},\;q =\tfrac{1}{3}


    Therefore: . P(\text{even}) \:=\:\frac{2}{3}

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