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    I'm stuck on this question: there are 7 red counters in a bag, the rest are blue. 2 counters were taken at random from the bag. The probability that here will be one of each colour is 7/15. What is the total number of colours in the bag?

    I have tried to form a probability tree and I came to the (wrong) conclusion, n being number of counters, that ((7/n) x ((n-7)/(n-1))) x (((n-7/n) x (7/(n-1))) = 7/15 - sorry I realise there are an unreal amount of brackets in this but I tried to make it as clear as possible... that kinda backfired lol
    If anyone could put me on the right track I would appreciate it! Or if you're feeling kind DM me the solution
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    (Original post by mundosinfin)
    I'm stuck on this question: there are 7 red counters in a bag, the rest are blue. 2 counters were taken at random from the bag. The probability that here will be one of each colour is 7/15. What is the total number of colours in the bag?

    I have tried to form a probability tree and I came to the (wrong) conclusion, n being number of counters, that ((7/n) x ((n-7)/(n-1))) x (((n-7/n) x (7/(n-1))) = 7/15 - sorry I realise there are an unreal amount of brackets in this but I tried to make it as clear as possible... that kinda backfired lol
    If anyone could put me on the right track I would appreciate it! Or if you're feeling kind DM me the solution
    Very close - just one small slip. Should be:

     \displaystyle \frac{7}{n}\times\frac{n-7}{n-1} +\frac{n-7}{n}\times\frac{7}{n-1}= \frac{7}{15}

    I trust you understand why.

    I get two possible values for n:
    Spoiler:
    Show


    10 or 21

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    (Original post by ghostwalker)
    Very close - just one small slip. Should be:

     \displaystyle \frac{7}{n}\times\frac{n-7}{n-1} +\frac{n-7}{n}\times\frac{7}{n-1}= \frac{7}{15}

    I trust you understand why.

    I get two possible values for n:
    Spoiler:
    Show


    10 or 21

    Ahhh that was careless haha, it’s good to see I was on the right track anyway. Thanks!
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