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Math Help - Dice -_-;

  1. #1
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    Dice -_-;

    Here's the question:

    In rolling a fair die, Find the probability of rolling a 1 before an even number? All I'm given as a hint is to consider

    P(1|1 or even)

    I know it's conditional probability, I just don't know if i need to use Bayes' theorem or something different, i know the answer is 1/4 but for some reason I keep coming up with 1/2, which cannot be right at all. I feel like I'm making a really dumb mistake and have been working on this one for hours.
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  2. #2
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    you are going to keep rolling the die until you get a 1 or an even number (since at that point, you know the answer).

    The chance of it being a 1 on that roll is P(1|1 or even)

    which is 1/4

    (since there are 4 values of the die that are 1 or even : 1,2,4,6)
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  3. #3
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    Hello, Dave2718!

    That hint was a good one . . .
    . . and SpringFan25 is absolutely correct.


    In rolling a fair die, find the probability of rolling a 1 before an even number?

    We can also solve this "the long way" . . .


    We are concerned with 3 events:

    . . \begin{array}{ccccc}P(\text{One}) &=& \frac{1}{6} \\ \\[-3mm]<br />
P(\text{Even}) &=& \frac{3}{6} \\ \\[-3mm] <br />
P(\text{Other}) &=& \frac{2}{6} \end{array}


    Then we have these possible outcomes:

    . . \begin{array}{cccccccc}<br />
P(\text{One on 1st roll}) &=& \frac{1}{6} \\ \\[-4mm]<br />
P(\text{One on 2nd roll}) &=& \left(\frac{2}{6}\right)\cdot\frac{1}{6} \\ \\[-4mm]<br />
P(\text{One on 3rd roll}) &=& \left(\frac{2}{6}\right)^2\cdot \frac{1}{6} \\ \\[-4mm]<br />
P(\text{One on 4th roll}) &=& \left(\frac{2}{6}\right)^3\cdot\frac{1}{6} \\ \\[-4mm]<br />
\vdots && \vdots \end{array}


    P(\text{One before Even}) \;=\;\frac{1}{6} + \frac{1}{6}\left(\frac{1}{3}\right) + \frac{1}{6}\left(\frac{1}{3}\right)^2 + \frac{1}{6}\left(\frac{1}{3}\right)^3 + \hdots

    . . . . . . . . . . . . . . . . =\;\frac{1}{6}\underbrace{\bigg[1 + \tfrac{1}{3} + \left(\tfrac{1}{3}\right)^2 + \left(\tfrac{1}{3}\right)^3 + \hdots \bigg]}_{\text{geometric series}}


    The sum of the geometric series is: . \dfrac{1}{1-\frac{1}{3}} \:=\:\dfrac{1}{\frac{2}{3}} \:=\:\frac{3}{2}


    Therefore: . P(\text{One before Even}) \;=\; \frac{1}{6}\cdot\frac{3}{2} \;=\;\boxed{\frac{1}{4}}

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