A bag contains 5 red marbles, 8 white marbles, 4 green marbles.
Question:

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

Solution:

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 :

$\mathrm{P}(\overline{\mathrm{C}})=1-\mathrm{P}(\mathrm{C})=1-\frac{4}{17}=\frac{13}{17}$

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