Three white and one red ball probability

Solution 1:

A winning: $$P(R_A)=\frac{1}{4}.$$ B winning: $$P(W_A)P(R_B|W_A)=\frac34 \cdot \frac13 = \frac14.$$ C winning: $$P(W_A)P(W_B|W_A)P(R_C|W_A\cap W_B)=\frac34 \cdot \frac23 \cdot \frac12=\frac14.$$

Solution 2:

Suppose that instead there are a black, a green, a blue and a red ball in the box. Three balls are taken out one by one without replacement.

Has the red ball a smaller or larger chance to be taken out at the third draw than e.g. the green ball? No, so the probability that it will be taken out at the third draw is $\frac14$.

The same reasoning tells us that this will also be true for the first or second draw.