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Math Help - ball probability

  1. #1
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    ball probability

    Bag A contains 3 red balls and 2 green balls whereas bag B contains 2 red balls and 3 green balls . Two balls are withdrawn from bag A to be put into bag B . Then two balls are taken out from B to be put in bag A . What is the probability that bag A now contains 3 red balls and 2 green balls ?
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  2. #2
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    Hello, thereddevils!

    Bag A contains 3 red balls and 2 green balls.
    Bag B contains 2 red balls and 3 green balls.
    Two balls are withdrawn from bag A to be put into bag B.
    Then two balls are taken out from bag B and put in bag A.
    What is the probability that bag A now contains 3 red balls and 2 green balls?
    After the exchange, bag A has been restored to its original state.
    Whatever colors were moved to bag B were subsequently returned to bag A.


    There are 3 cases to consider:

    . . [1]\;\overrightarrow{RR}\text{ then }\overleftarrow{RR}

    . . [2]\;\overrightarrow{RG}\text{ then }\overleftarrow{RG}

    . . [3]\;\overrightarrow{GG}\text{ then }\overleftarrow{GG}



    [1] Two Reds are moved both ways.

    . . P(\overrightarrow{RR}) \:=\:\frac{{3\choose2}}{{5\choose2}} \:=\:\frac{3}{10}

    . . Bag B now contains: 4 Red, 3 Green

    . . P(\overleftarrow{RR}) \:=\:\frac{{4\choose2}}{{7\choose2}} \:=\:\frac{6}{21}

    . . Hence: . P(\overrightarrow{RR} \wedge \overleftarrow{RR}) \:=\:\frac{3}{10}\cdot\frac{6}{21} \:=\:\frac{18}{210}


    [2] A Red and a Green are moved both ways.

    . . P(\overrightarrow{RG}) \:=\:\frac{{3\choose1}{2\choose1}}{{5\choose2}} \:=\:\frac{6}{10}

    . . Bag B now contains: 3 Red, 4 Green.

    . . P(\overleftarrow{RG}) \:=\:\frac{{3\choose1}{4\choose1}}{{7\choose2}} \:=\:\frac{12}{21}

    . . Hence: . P(\overrightarrow{RG} \wedge \overleftarrow{RG}) \:=\:\frac{6}{10}\cdot\frac{12}{21} \:=\:\frac{72}{210}


    [3] Two Greens are moved both ways.

    . . P(\overrightarrow{GG}) \:=\:\frac{{2\choose2}}{{5\choose2}} \:=\:\frac{1}{10}

    . . Bag B now contains: 2R, 5G.

    . . P(GG) \:=\:\frac{{5\choose2}}{{7\choose2}} \:=\:\frac{10}{21}

    . . Hence: . P(GG \wedge GG) \:=\:\frac{1}{10}\cdot\frac{10}{21} \:=\:\frac{10}{210}


    Therefore: . P(\text{bag A restored}) \;=\;\frac{18}{210} + \frac{72}{210} + \frac{10}{210} \;=\;\frac{100}{210} \;=\;\boxed{\frac{10}{21}}

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