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    this is the 3 mark question at the end of paper 1

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    after drawing out a tree diagram, I found the answer to be 6/11, is this correct?
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    (Original post by CraigBackner)
    this is the question at the end of paper 1

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    after drawing out a tree diagram, I found the answer to be 6/11, is this correct?
    That isn't what I got. Can you post your full working?
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    (Original post by sindyscape62)
    That isn't what I got. Can you post your full working?
    this is my tree diagram:
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    as you can see the probability of black in bag c is 6/11
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    (Original post by CraigBackner)
    this is my tree diagram:
    Name:  20170114_125833.jpg
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    as you can see the probability of black in bag c is 6/11
    You have drawn the tree diagram wrong (probabilities may be correct but I feel you're confused). In a tree diagram, the vertical sets of branches should represent the different events.

    So first you need to identify the events. The first event is taking a counter from A and putting it in B. And there are two outcomes for this : either a black is moved or a white is moved so you'll have two branches.

    What's the second event? Can you now draw a tree diagram? Try this and post your diagram.
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    (Original post by CraigBackner)
    this is my tree diagram:
    Name:  20170114_125833.jpg
Views: 143
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    as you can see the probability of black in bag c is 6/11
    Your tree diagram is correct, but you haven't quite understood what the question's asking. No balls are being taken out of bag c, the question is what's the chance that there are more black balls than white balls in c. Which branches of your tree diagram lead to more black than white balls in c?
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    (Original post by notnek)
    You have drawn the tree diagram wrong. In a tree diagram, the vertical sets of branches should represent the different events.

    So first you need to identify the events. The first event is taking a counter from A and putting it in B. And there are two outcomes for this : either a black is moved or a white is moved so you'll have two branches.

    What's the second event? Can you now draw a tree diagram? Try this and post your diagram.
    I'm confused, I've already accounted for the first event( which is moving a black or white counter from a to b) hence the probabilites of bag B change. Surely each event doesn't have to have a separate probability diagram?
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    (Original post by CraigBackner)
    I'm confused, I've already accounted for the first event( which is moving a black or white counter from a to b) hence the probabilites of bag B change. Surely each event doesn't have to have a separate probability diagram?
    You only need one diagram, but looking at yours made me feel like you didn't understand what was going on.

    The first set of two branches should be titled "Counter moved from A to B" or something like that. Similarly you should rename the second set of branches. And you only need two sets of branches because there are only two events.
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    (Original post by sindyscape62)
    Your tree diagram is correct, but you haven't quite understood what the question's asking. No balls are being taken out of bag c, the question is what's the chance that there are more black balls than white balls in c. Which branches of your tree diagram lead to more black than white balls in c?
    ohh i see! i thought it was essentially asking what is the probability of a black in bag c.
    So for me to acctually answer the question, do I just need to work out the probability of black and why by multiplying along the appropriate branches?
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    (Original post by CraigBackner)
    ohh i see! i thought it was essentially asking what is the probability of a black in bag c.
    So for me to acctually answer the question, do I just need to work out the probability of black and why by multiplying along the appropriate branches?
    Yes, that's it. There should be two branches that give you the result you're looking for.
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    (Original post by notnek)
    You only need one diagram, but looking at yours made me feel like you didn't understand what was going on.

    The first set of two branches should be titled "Counter moved from A to B" or something like that. Similarly you should rename the second set of branches. And you only need two sets of branches because there are only two events.
    ohh i see what you mean, and yeah it does make more sense to label it like that thanks! My tree diagram was setup to work out the new probabilities in each bag, thats why
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    (Original post by sindyscape62)
    Yes, that's it. There should be two branches that give you the result you're looking for.
    after mutilpying the branches i get the probability of black to be 288/1210 hence thats the answer?
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    (Original post by CraigBackner)
    ohh i see what you mean, and yeah it does make more sense to label it like that thanks! My tree diagram was setup to work out the new probabilities in each bag, thats why
    Even though your numbers were correct, your tree diagram didn't really make sense, which is why you got stuck.. Each set of branches should represent an event. "Bag A" as an event doesn't mean anything.
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    (Original post by CraigBackner)
    after mutilpying the branches i get the probability of black to be 288/1210 hence thats the answer?
    Which branches did you multiply?
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    (Original post by sindyscape62)
    Which branches did you multiply?
    i multiplied the probability of Black for each bag, a, b and c from my tree diagram , (BBB)
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    (Original post by CraigBackner)
    i multiplied the probability of B for each bag, a, b and c from my tree diagram , (BBB)
    You've still not understood the question. No balls are being removed from bag c so your last set of branching (which is about the probabilities of removing balls from c) isn't needed- I'd cross that column off your diagram.

    What you're looking for is situations where there are more black than white balls in c. What has to happen for this to be true?
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    for this to happen there has to be more black than white balls for a start, which there is.
    so for the proabability of black, you mutiply 6/10 by 8/11 which gives 48/110,whcih is the correct answer?
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    (Original post by CraigBackner)
    for this to happen there has to be more black than white balls for a start, which there is.
    so for the proabability of black, you mutiply 6/10 by 8/11 which gives 48/110,whcih is the correct answer?
    For there to be more black than white balls in c the ball moved from b to c must be black.

    There are two ways this can happen - WB or BB.

    The ball moved from a to b doesn't effect the number in c, so it can be either black or white. Since both of these branches give you the result you multiply through each one then add the two numbers.
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    (Original post by sindyscape62)
    For there to be more black than white balls in c the ball moved from b to c must be black.

    There are two ways this can happen - WB or BB.

    The ball moved from a to b doesn't effect the number in c, so it can be either black or white. Since both of these branches give you the result you multiply through each one then add the two numbers.
    Ohh i get it,it was my poorly made tree diagram which didnt allow me to understand it fully

    Thanks a lot for the help!
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    no its 76/110
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    no its 76/110
 
 
 
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