A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
(i) red
(ii) white
(iii) not green
Number of red marbles $=5$
Number of white marbles $=8$
Number of green marbles $=4$
Total number of marbles in the bag $=5+8+4=17$
$\therefore$ Total number outcomes $=17$
(i) Let $A$ be the event of drawing a red ball.
$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{5}{17}$
(ii) Let $\mathrm{B}$ be the event of drawing a white ball.
$\therefore \mathrm{P}(\mathrm{B})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{8}{17}$
(iii) Let $\mathrm{C}$ be the event of drawing a green ball.
$\therefore \mathrm{P}(\mathrm{C})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{4}{17}$
Now, the event of not drawing a green ball is :
$\mathrm{P}(\overline{\mathrm{C}})=1-\mathrm{P}(\mathrm{C})=1-\frac{4}{17}=\frac{13}{17}$