Six billiard balls, numbered $1$ through $6,$ are placed in a box. Three of the balls are red, and three are blue. One ball is to be drawn randomly from the box.
i. The probability that the ball drawn will be an even numbered red ball
ii. $\large \frac{1}{2}$
Question: Is Option i. greater than Option ii?
My attempt:
Probability of drawing $1$ red ball = $\large \frac{1}{2}.$
Probability of drawing a red ball AND a even number is $\large \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.$
Therefore my answer is no, since $\large \frac{1}{4}<\frac{1}{2}.$