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Math Help - Probability distribution--should these equal one?

  1. #1
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    Probability distribution--should these equal one?

    Hi there,

    I'm supposed to prepare a probability distribution for the following question:

    Two balls are drawn from a bag in which there are 4 white balls and 2 black balls. The number of black balls is being counted.

    Okay.

    So x = number of black balls.

    Doing combinations:


    Should those equal 1, or is it okay that they don't? I got .088 + .263 + .329, which doesnt = 1. Are my calculations correct?

    Thanks!!
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  2. #2
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    Hello, gobbajeezalus!

    I'm supposed to prepare a probability distribution for the following question:

    Two balls are drawn from a bag which has 4 white balls and 2 black balls.
    The number of black balls is being counted.


    0 Black: . \displaystyle {6\choose0}\!\left(\frac{1}{3}\right)^0\!\left(\fr  ac{2}{3}\right)^6

    1 Black: . \displaystyle {6\choose1}\left(\frac{1}{3}\right)^1\left(\frac{2  }{3}\right)^5

    2 Black: . \displaystyle {6\choose2}\left(\frac{1}{3}\right)^2\left(\frac{2  }{3}\right)^4


    Are my calculations correct? . Sorry, no . . .

    Your use of the "6" indicates that you are drawing all six balls from the bag.

    Your use of \tfrac{1}{3} and \tfrac{2}{3} repeatedly indicates that
    . . you are drawing the six balls with replacement.



    I will assume the two balls are drawn without replacement.

    \displaystyle \text{Then there are: }\;{6\choose2} \:=\:15\text{ possible outcomes.}


    \displaystyle P(\text{0B, 2W}) \;=\;\frac{{2\choose0}{4\choose2}}{15} \;=\;\frac{6}{15}

    \displaystyle P(\text{1B, 1W}) \;=\;\frac{{2\choose2}{4\choose1}}{15} \;=\;\frac{8}{15}

    \displaystyle P(\text{2B, 0W}) \;=\;\frac{{2\choose2}{4\choose0}}{15} \;=\;\frac{1}{15}
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