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    i spent a lot of time trying to figure this out but failed, thx in advance for the help

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    (Original post by Unicornman)
    i spent a lot of time trying to figure this out but failed, thx in advance for the help

    When it says that at least one of the counters is blue, that means that the probability that exactly one blue counter + probability that exactly two counters are blue are pulled out is 48/90.

    So, if there are x blue counters, then we know that:

    Then how can you write this probability down in terms of x?

    Remember that:
    P(exactly one blue counter) = P(blue counter)P(red counter) + P(red counter)P(blue counter)
    P(2 blue counters) = P(blue counter)P(blue counter).

    Also remember that they are not replaced. So, once the first blue counter is removed, the probability that the second counter is blah will have a different denominator.
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    (Original post by Zacken)
    When it says that at least one of the counters is blue, that means that the probability that exactly one blue counter + probability that exactly two counters are blue are pulled out is 48/90.

    So, if there are x blue counters, then we know that:

    Then how can you write this probability down in terms of x?

    Remember that:
    P(exactly one blue counter) = P(blue counter)P(red counter) + P(red counter)P(blue counter)
    P(2 blue counters) = P(blue counter)P(blue counter).

    Also remember that they are not replaced. So, once the first blue counter is removed, the probability that the second counter is blah will have a different denominator.
    Before I share my working with OP, I want to check I have worked this out correctly. I got 4 blue counters and 6 red ones.
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    (Original post by Wolfram Alpha)
    Before I share my working with OP, I want to check I have worked this out correctly. I got 4 blue counters and 6 red ones.
    thank you very much!
 
 
 
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