Question

7.
A bag contains 5 green balls and 7 blue balls. A ball is drawn at random from the bag. What is the
probability that the ball drawn is
(1) green or
(ii) not green​

Answer 1

Answer:

number of green balls=5

number of blue balls=7

total number of balls=12

probability of green balls =number of green balls/ total number of balls

=5/12

probability of not green balls =number of not green balls/ total number of balls

=7/12

Answer 2

Given:

Number of green balls=5

Number of blue balls=7

To find:

The probability of drawing a green ball and not a green ball

Solution:

The required probability is 5/12 and 7/12.

We can obtain the probability by dividing the favorable outcomes by the total possible outcomes.

The total balls=green balls+blue balls

=5+7=12 balls

Now, the total green balls=5 and blue balls=7.

i. The required probability of getting a green ball=number of green balls/total balls

Using the values,

=5/12

ii. Similarly, the required probability of getting not green balls=number of not green balls/total balls

=number of blue balls/total balls

=7/12

Therefore, the required probability is 5/12 and 7/12.