Question

A bag contains 4 red and 6 black balls. A ball is drawn at random, its colour is noted and it is returned to the bag. Moreover, 2 additional balls of the colour drawn are put in the bag and then a ball is drawn at random. What is the probability that the second ball is red?

• A. \( \frac{1}{3} \)
• B. \( \frac{2}{5} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{4}{5} \)

Hint:

Whenever the result of the second experiment depends on the outcome of the first experiment, use the Law of Total Probability: \[ P(A) = \sum P(B_i)P(A|B_i) \] Break the problem into cases based on the first event and then combine the probabilities.

Answer 1

The Correct Option is B

Concept: This problem uses the Law of Total Probability. If an event can occur through multiple mutually exclusive cases, then the total probability is the sum of probabilities through each case. \[ P(A) = P(B_1)P(A|B_1) + P(B_2)P(A|B_2) \] where \(B_1, B_2\) are different possible outcomes of the first event. Here,
• \(R_1\): First ball drawn is red
• \(B_1\): First ball drawn is black
• \(R_2\): Second ball drawn is red Thus, \[ P(R_2) = P(R_1)P(R_2|R_1) + P(B_1)P(R_2|B_1) \]

Step 1: Find the probability that the first ball is red or black. Initially the bag contains: \[ 4 \text{ red}, \quad 6 \text{ black} \] Total balls \(=10\). \[ P(R_1) = \frac{4}{10}, \qquad P(B_1) = \frac{6}{10} \]

Step 2: Find the probability that the second ball is red if the first ball was red. If the first ball drawn is red:
• The ball is returned.
• Two additional red balls are added. New composition: \[ 6 \text{ red}, \quad 6 \text{ black} \] Total balls \(=12\). \[ P(R_2|R_1) = \frac{6}{12} \]

Step 3: Find the probability that the second ball is red if the first ball was black. If the first ball drawn is black:
• The ball is returned.
• Two additional black balls are added. New composition: \[ 4 \text{ red}, \quad 8 \text{ black} \] Total balls \(=12\). \[ P(R_2|B_1) = \frac{4}{12} \]

Step 4: Apply the law of total probability. \[ P(R_2) = P(R_1)P(R_2|R_1) + P(B_1)P(R_2|B_1) \] \[ P(R_2) = \left(\frac{4}{10} \times \frac{6}{12}\right) + \left(\frac{6}{10} \times \frac{4}{12}\right) \] \[ P(R_2) = \frac{24}{120} + \frac{24}{120} \] \[ P(R_2) = \frac{48}{120} = \frac{2}{5} \] \[ \boxed{P(R_2) = \frac{2}{5}} \]