Step 1: Understand the question and make a note of the given information.

NOTE: A bag contains a red ball, yellow ball and blue ball.

Manasa takes out a ball from the bag without looking into it.

So, total number of possible outcomes = 3

Step 2: Determine the number of favourable of the different balls.

She takes out any one of them.

ASSUME:

Let Y be the event 'the ball taken out is yellow,

B be the event 'the ball taken out is blue, and

R be the event 'the ball taken out is red'.

- The number of outcomes favourable to the event Y = 1 \because number\ of\ yellow\ balls=1

Similarly,

- The number of outcomes favourable to the event B = 1
- The number of outcomes favourable to the event R = 1

Step 3: Calculate the probability of the individual balls

FORMULA: P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}

(i) So, Probability that she takes a yellow ball P(Y) = \frac{1}{3}

Similarly,

(ii) Probability that she takes a red ball P(R) = \frac{1}{3}

(iii) Probability that she takes a blue ball P(B) = \frac{1}{3}