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: Understanding the Concept:
Probability of an event = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
Step 3: Detailed Explanation:
Total balls = 30. Blue balls = \(m\).
Part (i):
Number of balls not blue = \(30 - m\).
\[ P(\text{not blue}) = \frac{30 - m}{30} \]
Part (ii):
Probability of drawing a blue ball in the first case (\(P_1\)) = \(\frac{m}{30}\).
Now, 6 blue balls are added.
New total balls = \(30 + 6 = 36\).
New blue balls = \(m + 6\).
New probability of drawing a blue ball (\(P_2\)) = \(\frac{m + 6}{36}\).
Given condition: \(P_2 = \frac{5}{4} P_1\)
\[ \frac{m + 6}{36} = \frac{5}{4} \times \frac{m}{30} \]
\[ \frac{m + 6}{36} = \frac{5m}{120} \]
\[ \frac{m + 6}{36} = \frac{m}{24} \]
Cross-multiplying:
\[ 24(m + 6) = 36m \]
Divide both sides by 12:
\[ 2(m + 6) = 3m \]
\[ 2m + 12 = 3m \Rightarrow m = 12 \]
Step 4: Final Answer:
(i) Probability is \(\frac{30-m}{30}\), (ii) The value of \(m\) is 12.