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Thread: dice

  1. #1
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    dice

    Throw outcome
    1 lose 3.rs
    2 lose 2.rs
    3 lose 1.rs
    4 neither
    5 win 1.rs
    6 win 5.rs

    this table represents a outcome of a single dice when thrown ...now what is probability of having of having at least 5.rs at the end of 2 throws ..


    my ans is 1/9 is it correct
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  2. #2
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    I get five different pairs each gives at least 5rs.
    Have I missread the question?
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  3. #3
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    Hello, arunachalam.s!


    Edit: Once again, I've made a really stupid blunder . . . embarrassing!


    A die is thrown. These are the payoffs.

    . . . $\displaystyle \begin{array}{c|c}
    \text{Throw} & \text{Payoff (\$)} \\ \hline
    1 & \text{-}3 \\ 2 & \text{-}2 \\ 3 & \text{-}1 \\ 4 & 0 \\ 5 & +1 \\ 6 & +5 \end{array}$

    What is the probability of having of having at least $5 at the end of 2 throws?

    My answer is $\displaystyle \tfrac{1}{9}$ . Is it correct? . . . . no
    There are $\displaystyle 6^2 = 36$ outcomes for two rolls of a die.


    To win at least $5, we must:

    . . Roll a 4 and a 6 in some order: .$\displaystyle (4,6),(6,4)$ . . . two ways.

    . . Roll a 5 and a 6 in some order: .$\displaystyle (5,6), (6,5)$ . . . two ways.

    . . Roll a 6 and a 6: .$\displaystyle {\color{blue}(6,6)}$ . . . one way.

    Hence, there are five ways to have at least $5.


    Therefore: .$\displaystyle P(\text{at least \$5}) \:=\:{\color{blue}\frac{5}{36}}$

    Last edited by Soroban; Aug 21st 2009 at 07:33 AM.
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