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Math Help - probleming solving technique confirmation

  1. #1
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    probleming solving technique confirmation

    i was doing these questions for probability and stats and i was wondering why the answers using two different methods dont match.
    i'll illustrate with examples
    Tree Diagrams (with worked solutions & videos)

    in the last two questions they use the outcome approach but if i attempt them using the probabilities of the branches i get the same answers.

    here's another question that i shortened just to illustrate


    Bag A contains 10 marbles of which 2 are read and 8 are black. Bag B contains 12 marbles of which 4 are red and 8 are black. A marble is drawn at random from each bag. Find the probability that both are red.


    for the above question using the outcome approach when the diagram is limited to 2 draws only.. i.e is a total of 4 outcomes.. my answer is 0.25
    but if i used the probabilities listed on the tree diagram my answer would be 2/10 * 4/12  = 1/15
    and these so dont match.

    so um.. what am i missing. could someone clarify the situation.
    is it because all the numbers or objects are not used here?
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  2. #2
    Senior Member Twig's Avatar
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    Quote Originally Posted by frixis View Post

    Bag A contains 10 marbles of which 2 are read and 8 are black. Bag B contains 12 marbles of which 4 are red and 8 are black. A marble is drawn at random from each bag. Find the probability that both are red.
    Hi!

    You have two different bags. You are to draw a ball from each of them. When you draw a ball from Bag A, it doesnt in any way affect the balls in Bag B. The probability of drawing a red ball from Bag A is independent of what you have drawn from Bag B and vice versa.

    Hence, the only way you can get two red balls is when you draw a red ball from Bag A and then from bag B (order is irrelevant).

    P(\mbox{ "Both balls are red" }) = \frac{\mbox{ number of red balls in bag A }}{\mbox{ total number of balls in bag A }} \cdot \frac{\mbox{ number of red balls in bag B }}{\mbox{ total number of balls in bag B }} = \frac{2}{10} \cdot \frac{4}{12}=\frac{1}{15}
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