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Thread: Losing my marbles.... Probability

  1. #1
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    Losing my marbles.... Probability

    OK so I have a bag of marbles containing: 3 Catseyes, 4 Agates and 7 Oxbloods. If all have an equal chance of being picked and two are drawn, what are the chances of at least 1 of the chosen being an Agate.
    According to my limited learning I worked it out as follows:

    $\displaystyle P(at least 1 agate)= \frac{4}{14} + \frac{4}{13} = \frac{54}{91} $
    Is this correct? If not then how do I work it out?
    Last edited by TommyBoy22; May 29th 2009 at 07:06 AM.
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  2. #2
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    Quote Originally Posted by TommyBoy22 View Post
    OK so I have a bag of marbles containing: 3 Catseyes, 4 Agates and 7 Oxbloods. If all have an equal chance of being picked and two are drawn, what are the chances of at least 1 of the chosen being an Agate.
    According to my limited learning I worked it out as follows:

    $\displaystyle P(at least 1 agate)= \frac{4}{14} + \frac{4}{13} = \frac{54}{91} $
    Is this correct? If not then how do I work it out?
    Hi TommyBoy,

    $\displaystyle P(\text{at least 1 agate})=1-(\frac{10}{14})(\frac{9}{13})$
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  3. #3
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    Hello, TommyBoy22!

    Sorry, your work is way off . . .
    What formulas (if any) are you using?


    I have a bag of marbles containing: 3 Catseyes, 4 Agates and 7 Oxbloods.
    If all have an equal chance of being picked and two are drawn,
    what is the probability of at least one Agate being chosen?
    There are 14 marbles: 3 CEs, 4 AGs, 7 OBs.

    Two marbles are drawn.
    . . There are: .$\displaystyle _{14}C_2 \:=\:91$ possible outcomes.

    The opposite of "at least one AG" is "no AGs."

    There are: 4 AGs and 10 Others.
    There are: .$\displaystyle _{10}C_2 \:=\:45$ ways to draw two Others (no AGs).
    . . Hence, there are: .$\displaystyle 91 - 45 \:=\:46$ ways to draw at least one AG.

    Therefore: .$\displaystyle P(\text{at least one AG}) \:=\:\frac{46}{91}$


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


    From your work, I assume you are working with a sequence of draws.
    In that case, we must consider all the cases.

    We want at least one AG.
    . . This means: (1 AG and 1 Other), or (2 AGs).


    There are two ways to get one AG and one Other:

    . . AG, then Other: .$\displaystyle \frac{4}{14}\cdot\frac{10}{13} \:=\:\frac{20}{91}$

    . . Other, then AG: .$\displaystyle \frac{10}{14}\cdot\frac{4}{13} \:=\:\frac{20}{91}$


    There is one way to get two AGs:

    . . AG, then AG: .$\displaystyle \frac{4}{14}\cdot\frac{3}{13} \:=\:\frac{6}{91}$


    Therefore: .$\displaystyle P(\text{one AG or two AGs}) \;=\;\frac{20}{91} + \frac{20}{91} + \frac{6}{91} \;=\;\frac{46}{91}$

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  4. #4
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    Hi thanks guys.. yeah no theory used.. just made sense in my head
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