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