**Question:**

A box contains 5 red marbels, 8 white marbles and 4 green marbles, One marble is taken out of the box at ramdom. What is the probability that the marble taken out will be

(i) red?

(ii) white?

(iii) not green?

**Solution:**

GIVEN: A box contains 5 red, 4 green and 8 white marbles and a marble is drawn at random

TO FIND: Probability of getting a marble

(i) red

(ii) white

(iii) not green

Total number of marble:

(i) Total number red marble are 5

We know that PROBABILITY =

Hence probability of getting red marble

(ii) Total number of white marbles are 8

We know that PROBABILITY =

Hence probability of getting white marble is

(iii) Total number of green marbles is 4

We know that PROBABILITY =

Hence probability of getting green marbles

We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.

$P(E)+P(\bar{E})=1$

$\frac{4}{17}+P(\bar{E})=1$

$P(\bar{E})=1-\frac{4}{17}$

$P(\bar{E})=\frac{17-4}{27}=\frac{13}{17}$

Hence the probability of getting green marbles is $P(\bar{E})=\frac{13}{17}$