# Consider a box containing four balls: one red, one green, one blue, and one tricolor (=red,green and blue)

Consider a box containing four balls: one red, one green, one blue, and one tricolor (=red,green and blue).

You draw one ball from the box.

Consider the three events:

R = {the drawn ball contains red}

G = {the drawn ball contains green}

Y = {the drawn ball contains red and green}

Are R and G independent? At first I thought yes, but doesn't getting a tricolor ball affect the probability of R and G?

Are G and Y independent? I think no because a tricolor ball can be in both G and Y.

Without any extra information, R and G are both $\frac{2}{4} = \frac{1}{2}$. Now suppose I know that R is true. The ball can now be red or tri-color. That means if R is true, G is true $\frac{1}{2}$ of the time. The probability didn't change, so G is independent of R, and the same argument can be applied vice-versa.
Using conditional probability formulas: $$P(R|G)=\frac{P(R\cap G)}{P(G)}=\frac{1/4}{2/4}=\frac12=P(R) \Rightarrow Independent;$$ $$P(R|Y)=\frac{P(R\cap Y)}{P(Y)}=\frac{1/4}{1/4}=1\ne \frac12=P(R) \Rightarrow Dependent.$$