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Math Help - Basic Prob

  1. #1
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    Basic Prob

    please find attached file as below:

    i have try by using tree diagram until the part which i mark with BOX.other than that i have blur how to configure it..any advise

    tq
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  2. #2
    Member rtblue's Avatar
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    There are a total of 9 red discs and 7 black disks. The bags have been mixed thoroughly, so it is almost as if the contents have been dumped onto the ground and mixed. After mixing, there are 9 discs in the first bag. The probability of there being exactly three is:

    \frac{9}{16}*\frac{8}{16}*\frac{7}{16}*\frac{7}{16  }*\frac{6}{16}*\frac{5}{16}*\frac{4}{16}*\frac{3}{  16}*\frac{2}{16}

    The first three terms account for there being 3 red discs, and the last six terms account for there being 6 black discs.
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  3. #3
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    Dear rtblue..

    tq for the post.

    attached is the ans:
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  4. #4
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    Hello, nikk!

    \text{Bag 1 contains 3 red and 5 black discs.}
    \text{Bag 2 contains 6 red and 2 black discs.}

    \text{A disc is chosen at random from Bag 1 and placed in Bag 2.}
    \text{Then a disc is chosen at rancom from Bag 2 and placed in Bag 1.}

    \text{Find the probability that Bag 1 has 3 red discs.}

    We want the probability that Bag 1 ends up with its original contents.


    There are two ways this can happen:
    . . [1] A red is drawn from Bag 1 and a red is returned to Bag 1.
    . . [2] A black is drawn from Bag 1 and a black is returned to Bag 1.


    [1] A red is drawn from Bag 1: . P(R_1) = \frac{3}{8}
    . . .Bag 2 now contains:7 red and 2 black discs:
    . . .A red is drawn from Bag 2: . P(R_2) \:=\:\frac{7}{9}
    . . . . . P(R_1R_2) \:=\:\frac{3}{8}\cdot \frac{7}{9}

    [2] A black is drawn from Bag 1: P(B_1) = \frac{5}{8}
    . . .Bag 2 now contains: 6 red and 3 black discs.
    . . .A black is drawn from Bag 2: . P(B_2) = \frac{3}{9}
    . . . . . P(B_1B_2) \:=\:\frac{5}{8}\cdot\frac{3}{9}


    Therefore: . P(R_1R_2\text{ or }R_1B_2) \;=\;\frac{3}{8}\cdot\frac{7}{9} + \frac{5}{8}\cdot\frac{3}{9} \;=\;\frac{7}{24} + \frac{5}{24} \;=\;\frac{12}{24} \;=\;\dfrac{1}{2}

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