# probability of picking up two m&ms of same color randomly

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.

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

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 $\frac{3}{16} *\frac{2}{15}+ \frac{5}{16}* \frac{4}{15}+ \frac{1}{2}* \frac{7}{15} = \frac{82}{240}= \frac{41}{120}$