**Question:**

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

**Solution:**

if a ball is drawn out of 22 balls (5 blue + 7 green + 10 red), then the total number of outcomes are

$n(S)=22$

(i) Let $E_{1}=$ Event of getting a red ball

$n\left(E_{1}\right)=10$

$\therefore \quad$ Required probability $=\frac{n\left(E_{1}\right)}{n(S)}=\frac{10}{22}=\frac{5}{11}$

(ii) Let $E_{2}=$ Event of getting a green ball

$n\left(E_{2}\right)=7$

$\therefore \quad$ Required probability $=\frac{n\left(E_{2}\right)}{n(S)}=\frac{7}{22}$

(iii) Let $E_{3}=$ Event getting a red ball or a green ball $i, e$., not a blue ball.

$n\left(E_{3}\right)=(10+7)=17$

$\therefore \quad$ Required probability $=\frac{n\left(E_{3}\right)}{n(S)}=\frac{17}{22}$