A bag contains 30 balls out of which 'm' number of balls are blue in colour. a) Find the probability that a ball drawn at random from the bag is not blue. b) If 6 more blue balls are added in the bag,

Here's the solution for question 29: 29. A bag contains 30 balls out of which 'm' number of balls are blue in colour. a) Find the probability that a ball drawn at random from the bag is not blue. Step 1: Identify the total number of balls and the number of blue balls. Total number of balls = 30 Number of blue balls = m Step 2: Calculate the number of balls that are not blue. Number of non-blue balls = Total balls - Number of blue balls = 30 - m Step 3: Calculate the probability of drawing a ball that is not blue. P(not blue) = Number of non-blue ballsTotal number of balls P(not blue) = (30 - m)/(30) The probability that a ball drawn is not blue is (30 - m)/(30). b) 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. Step 1: Calculate the probability of drawing a blue ball in the first case (P_1(blue)). P_1(blue) = Number of blue ballsTotal number of balls = (m)/(30) Step 2: Determine the new number of blue balls and total balls after adding 6 blue balls. New number of blue balls = m + 6 New total number of balls = 30 + 6 = 36 Step 3: Calculate the probability of drawing a blue ball in the new case (P_2(blue)). P_2(blue) = New number of blue ballsNew total number of balls = (m+6)/(36) Step 4: Set up the equation based on the given condition. The problem states that P_2(blue) = (5)/(4) × P_1(blue). (m+6)/(36) = (5)/(4) × (m)/(30) Step 5: Simplify and solve for m. (m+6)/(36) = (5m)/(120) Simplify the right side by dividing 5 and 120 by 5: (m+6)/(36) = (m)/(24) Cross-multiply: 24(m+6) = 36m 24m + 144 = 36m Subtract 24m from both sides: 144 = 36m - 24m 144 = 12m Divide by 12: m = (144)/(12) m = 12 The value of m is 12. That's 2 down. 3 left today — send the next one.