Question

A bag contains 30 balls out of which 'm' number of balls are blue in colour.
(i) Find the probability that a ball drawn at random from the bag is not blue.
(ii) If 6 more blue balls are added in the bag, then the probability of drawing a blue ball will be \(\frac{5}{4}\) times the probability of drawing a blue ball in the first case. Find the value of m.

Hint:

Remember that adding items to a bag changes both the specific favorable count AND the total outcome count. Always update the denominator!

Answer 1

Step 1: Total Balls and Given Data
Total balls = 30
Number of blue balls = m

Part (i): Probability of Not Blue
Number of balls that are not blue = 30 − m

Probability (not blue) = (30 − m) / 30

Part (ii): After Adding 6 Blue Balls
Initial probability of blue:
P₁ = m / 30

After adding 6 blue balls:
New total balls = 36
New blue balls = m + 6

New probability:
P₂ = (m + 6) / 36

Given:
P₂ = (5/4) P₁

So,
(m + 6) / 36 = (5/4)(m / 30)

Simplify RHS:
(5/4)(m / 30) = 5m / 120
= m / 24

Thus,
(m + 6) / 36 = m / 24

Cross multiply:
24(m + 6) = 36m

24m + 144 = 36m
144 = 12m
m = 12

Final Answer:
(i) Probability of not blue = (30 − m) / 30
(ii) Number of blue balls = 12