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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 \(0\).