Question

A box \( P \) contains one white ball, three red balls and two black balls. Another box \( Q \) contains two white balls, three red balls and four black balls. If one ball is drawn at random from each one of the two boxes, then the probability that the balls drawn are of different color is

• A. \( \frac{29}{54} \)
• B. \( \frac{25}{42} \)
• C. \( \frac{35}{54} \)
• D. \( \frac{39}{52} \)

Hint:

When calculating the probability of combined events from independent sources, consider each source separately and then integrate the probabilities to find the final result.

Answer 1

The Correct Option is A

Step 1: Identify the total number of balls in each box: - Box \( P \): 6 balls (1 white, 3 red, 2 black) - Box \( Q \): 9 balls (2 white, 3 red, 4 black) 
Step 2: Calculate the probabilities for each case where the colors differ: - Case 1: White from \( P \) and Non-white from \( Q \) \[ P = \frac{1}{6} \times \frac{7}{9} = \frac{7}{54} \] - Case 2: Red from \( P \) and Non-red from \( Q \) \[ P = \frac{3}{6} \times \frac{6}{9} = \frac{18}{54} \] - Case 3: Black from \( P \) and Non-black from \( Q \) \[ P = \frac{2}{6} \times \frac{5}{9} = \frac{10}{54} \] Step 3: Sum the probabilities: \[ P({different colors}) = \frac{7}{54} + \frac{18}{54} + \frac{10}{54} = \frac{35}{54} \]