A box contains 3 red, 8 yellow, and 13 green balls A box contains 3 red, 8 yellow, and 13 green balls.  Another box contains 5 red, 7 yellow, and 6 green balls. One ball is selected at random from each box. What is the probability that both balls will be of the same color.
I believe red has a probability of $3 \div 24 \times 5 \div 18$ but I am not sure.  Is this correct?  If so then do you just do the same for the other colors and add them all together?
Thank you.
 A: If the experiments are independent, you have three successful outcomes: RR, YY, GG. 
A: Let's look at it like this. My combinatorics professor in college always emphasized on the semantics of mathematical language. What does this mean? Well since both of your pulls are separate but at the same time. You can do Box 1 and Box 2. In mathematical sense 'and' is ='+'. Since you have three colors of balls with a probably for each box. You will want to combined each probability, or 1 or 2. In this sense "or" = '$\times$'. You combine the probabilities of each like so
$$
        (\frac{1}{8}\times\frac{5}{18})+(\frac{1}{3}\times\frac{7}{18})+(\frac{13}{24}\times\frac{1}{3})
$$
the result of which will be
$$
        \frac{149}{432} = .34491 = 35\%
$$
if someone could check my work, but I think this might be the right answer. If this answer is incorrect could someone please supplement me with a response that explains my error. Again my apologies if it is incorrect Probability is something I haven't touched in a couple years. 
A: The number of possible outcomes is $24\times18$.
The number of favorable outcomes are as follows:
$(3\times5)+(8\times7)+(13\times6)$
The probability is simply: $\frac{(3\times5)+(8\times7)+(13\times6)}{24\times18}=\frac{149}{432}$ 
