A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observ

Q: Wait, why does the total number of balls change to 12? Shouldn't it stay 10?

That is a classic trap! Read the problem carefully: 'this ball along with two additional balls of the same colour are returned to the bag.' You are returning the ball you drew (so the count stays the same) AND adding two *new* balls. So, $10 + 2 = 12$. Always track the total count after every modification!

Q: Can I just use symmetry and say the probability is $4/10$?

Intuition is a powerful tool, but in probability, it can be a dangerous trap. While symmetry works in some problems, here the 'reinforcement' (adding balls of the same color) biases the second draw. You must calculate the paths explicitly to see how the reinforcement changes the odds.

Q: Why do we multiply the probabilities in the formula?

We multiply because these are sequential events. The probability of a specific path (e.g., Red then Red) is the probability of the first step multiplied by the probability of the second step *given* the first occurred. It is the fundamental 'AND' rule in probability theory.

Question

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

Options:

(A) 3/4

(B) 3/10

(C) 2/5

(D) 1/5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a probability problem; we are detectives investigating a dynamic system.

Imagine you are standing before a bag containing $4$ red balls and $6$ black balls. The moment you reach in, the system changes, illustrating how information and actions alter the landscape of possibility.

### The First Branch

We begin with an initial state of $10$ balls. When you draw the first ball, you face two mutually exclusive paths.

The probability of drawing a red ball first is:
$$P(R_1) = \frac{4}{10}$$

Conversely, the probability of drawing a black ball first is:
$$P(B_1) = \frac{6}{10}$$

### The Twist in the Tale

The rule is to return the ball and add two more of the same color. This creates a feedback loop.

If we drew a red ball, the red count jumps from $4$ to $6$, and the total count rises from $10$ to $12$. If we drew a black ball, the black count jumps from $6$ to $8$, and the total count also rises to $12$.

### The Conditional Reality

We now calculate the conditional probabilities for the second draw. If we are in the "Red First" universe, the probability of drawing a red ball again is:
$$P(R_2|R_1) = \frac{6}{12}$$

If we are in the "Black First" universe, the probability of drawing a red ball is:
$$P(R_2|B_1) = \frac{4}{12}$$

### The Synthesis

To find the total probability of drawing a red ball in the second attempt, $P(R_2)$, we use the **Total Probability Theorem**. We sum the probabilities of all paths that lead to our desired outcome:

$$P(R_2) = P(R_1) \cdot P(R_2|R_1) + P(B_1) \cdot P(R_2|B_1)$$

Substituting our values, we get:

$$P(R_2) = \left( \frac{4}{10} 	imes \frac{6}{12} ight) + \left( \frac{6}{10} 	imes \frac{4}{12} ight)$$

Both terms result in $\frac{24}{120}$. Adding them together yields:

$$P(R_2) = \frac{24}{120} + \frac{24}{120} = \frac{48}{120}$$

### The Final Victory

Simplifying $\frac{48}{120}$ is the final step of our journey. Dividing both the numerator and the denominator by $24$, we arrive at our destination:

**Final Answer:**
$$P(R_2) = \frac{2}{5}$$

This problem teaches us that in complex systems, we must map out every possible path. By breaking the problem into clear, logical steps, you turn a daunting question into a simple, elegant calculation.

Select Answer:

Visualized Solution

Initial State of the Bag

The Two-Step Experiment

Case 1: First Ball is Red ( $R_{1}$ )

Updating the Bag (Case 1)

Probability of Red in Second Draw (Case 1)

Case 2: First Ball is Black ( $B_{1}$ )

Updating the Bag (Case 2)

Probability of Red in Second Draw (Case 2)

Applying Total Probability Theorem

Substitution of Values

Multiplication of Fractions

Final Addition and Simplification

Summary and Key Takeaway

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

The First Branch

The Twist in the Tale

The Conditional Reality

The Synthesis

The Final Victory

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