### Theory:

Let us understand few concepts using examples.
Example:
1. An urn contains a green ball, an orange ball, a white ball, and a red ball. What is the probability of choosing

(i) a green ball,  (ii) an orange ball,  (iii) a white ball,  (iv) a red ball.

Solution:

Let $$G$$ denote the green ball, $$O$$ denote the orange ball, $$W$$ denotes the white ball, and $$R$$ denotes the red ball.

Here, the number of possible outcomes $$= 4$$.

(i) $$P(G) = \frac{\text{Number of outcomes favourable to G}}{\text{Number of all possible outcomes}} = \frac{1}{4}$$

(ii) $$P(O) = \frac{1}{4}$$

(iii) $$P(W) = \frac{1}{4}$$

(iv) $$P(R) = \frac{1}{4}$$

Important!
In this example, observe that every event has only one outcome of the experiment. This type of event is called an elementary event.

Note that: $$P(G) + P(O) + P(W) + P(R) = 1$$

That is, the sum of the probabilities of all the elementary events of an experiment is $$1$$.

2. A coin is tossed once. (i) What is the probability of getting a head? (ii) What is the probability of getting a tail?

Solution:

Let $$H$$ denote the head, and $$T$$ denote the tail.

(i) $$P(H) = \frac{1}{2}$$

(ii) $$P(T) = \frac{1}{2}$$

Important!
Here, we observe that $$P(H) + P(T) = 1$$. Note that $$T$$ and 'not $$H$$' are the same. Let us denote 'not $$H$$' as $$\overline H$$.

Therefore, the above equation becomes $$P(H) + P(\overline H) = 1$$ where 'not $$H$$' is the complement of the event $$H$$.

Hence, we say that $$H$$ and $$\overline H$$ are complementary events.
If the probability of an event is $$0$$, then the event is said to be an impossible event.

If the probability of an event is $$1$$, then the event is said to be a sure event or a certain event.

3. A bag contains $$2$$ white balls and a yellow ball. What is the probability of getting a red ball?

Solution:

Let $$R$$ denote the red ball.

Here, the total number of balls in the bag $$= 3$$

Since the bag contains only white and yellow balls, there is no chance of getting an outcome of a red ball.

Thus, $$P(R) = \frac{0}{3} = 0$$

Therefore, the probability of getting a red ball is $$0$$.