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Can someone check my probability answers please watch

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    1. A jar contains 4 red, 6 white and 5 green discs. 3 are removed and are not replaced. Are these right?:

    p(1st is red) 4/15
    p(2nd is white if first is green) 1/7
    p(2 of the three discs are red) 0.149
    p(3 discs are the same colour) 0.104

    If I have gone wrong would you mind telling me how I correct it? Thanks
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    the first one is wrong anyway, I haven't checked the others.

    4 red 18 in total so the answer is 4/18 or 2/9
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    First two are correct.

    What's your working for the last 2 qns?
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    No there is 15 in total so p(1st is red) is 4/15???
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    oops sorry didn't read the question properly and added the 3 as well. Sorry:confused:
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    I think the second one is 6/14 (3/7). I know that you've multiplied the fractions to get 1 seventh, but they've already told you that green is the first one - so you just take the probability of white, 6/15, and minus 1 from the denominator because one green has been taken.
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    I thought it might be that. These old papers don't have the 'given' word on it so I don't look at it in the same way lol

    Thanks
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    Also, how did you get that for the last one?
    Wouldn't it be:
    (\frac{4}{15} \times\frac{3}{14} \times\frac{2}{13})+(\frac{6}{15  } \times\frac{5}{14} \times\frac{4}{13})+(\frac{5}{15  } \times\frac{4}{14} \times\frac{3}{13})=0.0747
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    Hmm yes, thank you and the 3rd?
 
 
 
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Updated: March 26, 2011
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