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Math Help - Confused about probability and Tree diagrams

  1. #1
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    Confused about probability and Tree diagrams

    The problem is
    A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls. Two balls are chosen from the jar, with replacement. What is the probability that you choose a red and a yellow ball?

    I know that they do not affect each other so could be independent events, so P(red and yellow) = P(red) x P(yellow) so 6/21 x 1/3 = 2/21

    However, I do not know why my tree diagram isn't working

    ----6/21---- Red
    ---6/21--- Red
    ---7/21---Yellow
    ---3/21---Green
    ---5/21---White
    ----7/21---- Yellow
    ---6/21--- Red
    ---7/21---Yellow
    ---3/21---Green
    ---5/21---White
    ----3/21---- Green
    ---6/21--- Red
    ---7/21---Yellow
    ---3/21---Green
    ---5/21---White
    ----5/21---- White
    ---6/21--- Red
    ---7/21---Yellow
    ---3/21---Green
    ---5/21---White
    I've been taught you multiply across for the probability.
    I get two values for the probability.
    P(yellow and red)=2/21
    P(red and yellow)=2/21

    Do you not add these two together to get 4/21?

    I'm really confused now as to which one is correct.
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  2. #2
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    The order matters as replacement is allowed.

    Choosing a red and a yellow can happen in 2 ways....

    First a red is chosen, then a yellow.
    Or, first a yellow is chosen, then a red.

    P(red and yellow) is just one of these.
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  3. #3
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    Hi Henryblah,
    The addition rule applies
    P of red or yellow first draw = 13/21
    P of red or yellow not being drawn on second draw = 9/21
    P of red and yellow two draws = 13/21 - 9/21 = 4/21
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  4. #4
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    Hello, Henryblah!

    A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls.
    Two balls are chosen from the jar, with replacement.
    What is the probability that you choose a red and a yellow ball?
    I assume this means "in any order?".

    I know that they do not affect each other so they are independent events.
    So: P(\text{red \& yellow}) \,=\, P(\text{red}) \times P(\text{yellow}) \,=\,\frac{6}{21} \times \frac{1}{3} \:=\: \frac{2}{21} . Right!


    However, I do not know why my tree diagram isn't working.
    You made it complicated, but it is correct.


    \begin{array}{ccc}\text{Red: } \frac{6}{21} \\<br />
& \text{Yellow:} & \frac{7}{21} \\ \\[-4mm]<br />
& \text{Other:} & \frac{14}{21} \end{array}


    \begin{array}{ccc}\text{Yellow: }\:\frac{7}{21} \\<br />
& \text{Red:} & \frac{6}{21} \\ \\[-4mm]<br />
& \text{Other:} & \frac{15}{21} \end{array}


    I've been taught you multiply across for the probability. .Correct!

    I get two values for the probability.
    . . \begin{array}{c}P(\text{Yellow, then Red})\,=\,\frac{2}{21} \\ \\[-4mm]<br />
P(\text{Red, then Yellow})\,=\,\frac{2}{21} \end{array}

    Do you not add these two together to get \dfrac{4}{21}? . Yes!

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  5. #5
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    Thank you. I think I understand this now.

    Also @ whoever put my thread here
    sorry for putting this in the wrong section. I didn't see there was a probability forum.
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