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

Step 1: Understanding the Question:
The problem describes a two-stage experiment where the outcome of the second draw depends on the result of the first draw.
We need to find the total probability of drawing a red ball in the second attempt, considering two mutually exclusive cases: the first ball was red or the first ball was black.
Step 2: Key Formula or Approach:
We use the Law of Total Probability:
\[ P(R_2) = P(R_1) \cdot P(R_2|R_1) + P(B_1) \cdot P(R_2|B_1) \]
where:
\( R_1, B_1 \) are events of drawing a red or black ball first.
\( R_2 \) is the event of drawing a red ball second.
Step 3: Detailed Explanation:
Initial composition: 4 Red (R) and 6 Black (B). Total = 10 balls.
Probabilities of the first draw:
\[ P(R_1) = \frac{4}{10}, \quad P(B_1) = \frac{6}{10} \]
Case 1: First ball is Red (\( R_1 \))
The ball is returned, and 2 more red balls are added.
New composition: \( 4 + 2 = 6 \) Red, 6 Black. Total = 12 balls.
\[ P(R_2|R_1) = \frac{6}{12} \]
Case 2: First ball is Black (\( B_1 \))
The ball is returned, and 2 more black balls are added.
New composition: 4 Red, \( 6 + 2 = 8 \) Black. Total = 12 balls.
\[ P(R_2|B_1) = \frac{4}{12} \]
Total Probability calculation:
\[ P(R_2) = \left( \frac{4}{10} \times \frac{6}{12} \right) + \left( \frac{6}{10} \times \frac{4}{12} \right) \]
\[ P(R_2) = \frac{24}{120} + \frac{24}{120} = \frac{48}{120} \]
Simplifying the fraction:
\[ P(R_2) = \frac{2}{5} \]
Step 4: Final Answer:
The probability that the second ball drawn is red is \( \frac{2}{5} \).