$A$ bag contains $10$ red,$5$ blue and $7$ green balls. $A$ ball is drawn at random. Find the probability of this ball being a
$(i)$ red ball
$(ii)$ green ball
$(iii)$ not a blue ball

(N/A) The total number of balls in the bag is $10 + 5 + 7 = 22$.
Thus,the total number of possible outcomes is $n(S) = 22$.
$(i)$ Let $E_1$ be the event of drawing a red ball. The number of red balls is $n(E_1) = 10$.
Therefore,the probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{10}{22} = \frac{5}{11}$.
$(ii)$ Let $E_2$ be the event of drawing a green ball. The number of green balls is $n(E_2) = 7$.
Therefore,the probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{7}{22}$.
$(iii)$ Let $E_3$ be the event of not drawing a blue ball. This means drawing either a red or a green ball.
The number of favorable outcomes is $n(E_3) = 10 + 7 = 17$.
Therefore,the probability $P(E_3) = \frac{n(E_3)}{n(S)} = \frac{17}{22}$.