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There are 3 red m&ms, 5 green m&ms, and 8 blue m&ms. If I pick two m&ms out randomly, what is the probability of me picking two m&ms of the same color? I'm not sure if this is correct but I think it's $2/15 + 4/15 + 7/15$ as those are all the separate chances of a color being selected again added together, but I wonder if I might be extra counting...Just wondering if that's correct.

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  • $\begingroup$ One way: The probability of two red is $\frac{3}{16}\cdot\frac{2}{15}$. Similar expressions for two green, two blue, add. $\endgroup$ May 2, 2015 at 20:12

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You would be correct if you had an equal probability of choosing any color. You have a $ \frac{3}{16} $ chance of picking a red m&m, a $ \frac{5}{16} $ chance of picking a green m&m, and a $ \frac{1}{2} $ chance of picking a blue m&m. You are correct in your logic of the probabilities, so that makes the total probability of picking two m&ms randomly $ \frac{3}{16} *\frac{2}{15}+ \frac{5}{16}* \frac{4}{15}+ \frac{1}{2}* \frac{7}{15} = \frac{82}{240}= \frac{41}{120} $

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No. If you first pick a red one (3/16), the probability of picking another red one is 2/15, giving 6/240

If the first is green (5/16), the probability of picking another green one is 4/15, giving 20/240

If the first is blue (8/16), the probability of picking another blue one is 7/15, giving 56/240.

So the total is 82/240 = 41/120, which is not 13/15.

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  • $\begingroup$ According to the OP there are 16 m&ms in total $\endgroup$
    – Ynhockey
    May 2, 2015 at 20:15

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