Krishna
0

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}