Question

A bag contains 30 balls out of which ‘m’ number of balls are blue in colour.

1. Find the probability that a ball drawn at random from the bag is not blue.
2. If 6 more blue balls are added in the bag, then the probability of drawing a blue ball will be `5/4` times the probability of drawing a blue ball in the first case. Find the value of m.

Answer 1

Given:

Total balls = 30.

Blue balls = m.

i. Number of balls not blue = 30 – m.

Probability of an event = `"Number of favorable outcomes"/"Total number of outcomes"`

P(not blue) = `(30 - m)/30`

ii. Probability of drawing a blue ball in the first case `(P_1) = m/30`.

Now, 6 blue balls are added.

New total balls = 30 + 6 = 36.

New blue balls = m + 6.

New probability of drawing a blue ball `(P_2) = (m + 6)/36`.

Given condition: `P_2 = 5/4 P_1`

`(m + 6)/36 = 5/4 xx m/30`

`(m + 6)/36 = (5m)/120`

`(m + 6)/36 = m/24`

24(m + 6) = 36m

2(m + 6) = 3m

2m + 12 = 3m

⇒ m = 12