Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn

Q: Why can't I just use the initial probabilities? Why do I need to consider the transfers?

Because the transfers are not independent events. The second transfer depends on the outcome of the first, and the final draw depends on the outcome of the second. You must map the entire history to get the correct result.

Q: Is there a shortcut using symmetry?

While symmetry is a powerful tool in physics, here the urns start with different compositions. Urn A has $6$ Red, Urn B has $4$ Red. This asymmetry means you cannot simply average the results. You must calculate each path individually.

Q: What if I get confused with the denominators?

Always keep track of the total number of balls. After the first transfer, Urn B has $11$ balls. After the second transfer, Urn A returns to $10$ balls. Keeping the denominator consistent in your head will prevent calculation errors.

Question

Urn $A$ contains 6 red and 4 black balls and urn $B$ contains 4 red and 6 black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is found to be red is .........

Sigma Percentile · Accepted Answer

### The Anatomy of the Urn Problem

Welcome, my dear student. Today, we are going to peel back the layers of a classic probability puzzle.

Imagine you are standing before two urns, Urn A and Urn B. Urn A is a vessel of $6$ red and $4$ black balls, while Urn B holds $4$ red and $6$ black balls.

Our goal is to find the probability of drawing a red ball from Urn A after a two-stage transfer process ($A 	o B$ then $B 	o A$).

### The Four Paths of Destiny

Because the balls transferred can be either red or black, we must account for every possible history. We have four mutually exclusive paths to consider.

**Case 1: Red to Red**

We transfer a red ball from A to B, and then a red ball from B back to A. The probability of the first transfer is $\frac{6}{10}$.

After the transfer, Urn B has $5$ red balls out of $11$. The return transfer probability is $\frac{5}{11}$, and Urn A is left with $6$ red balls out of $10$.

$$P_1 = \frac{6}{10} 	imes \frac{5}{11} 	imes \frac{6}{10} = \frac{180}{1100}$$

**Case 2: Red to Black**

We transfer a red ball from A to B, but a black ball from B to A. The first transfer is $\frac{6}{10}$.

Urn B now has $6$ black balls out of $11$, so the return transfer is $\frac{6}{11}$. Urn A lost a red and gained a black, leaving it with $5$ red balls out of $10$.

$$P_2 = \frac{6}{10} 	imes \frac{6}{11} 	imes \frac{5}{10} = \frac{180}{1100}$$

**Case 3: Black to Red**

We transfer a black ball from A to B, and a red ball from B to A. The probability of drawing a black ball from A is $\frac{4}{10}$.

Urn B receives this black ball, and we draw one of its $4$ red balls to send back, which is $\frac{4}{11}$. Urn A now has an extra red ball, making it $7$ red out of $10$.

$$P_3 = \frac{4}{10} 	imes \frac{4}{11} 	imes \frac{7}{10} = \frac{112}{1100}$$

**Case 4: Black to Black**

Finally, a black ball goes from A to B, and a black ball returns from B to A. The first transfer is $\frac{4}{10}$.

Urn B now has $7$ black balls, so the return transfer is $\frac{7}{11}$. Urn A gets a black ball back, so its red ball count remains $6$ out of $10$.

$$P_4 = \frac{4}{10} 	imes \frac{7}{11} 	imes \frac{6}{10} = \frac{168}{1100}$$

### The Synthesis

Now, we invoke the **Law of Total Probability**. We must sum these four mutually exclusive paths to find the overall probability.

$$P(	ext{Total}) = P_1 + P_2 + P_3 + P_4 = \frac{180 + 180 + 112 + 168}{1100} = \frac{640}{1100}$$

Simplifying this, we cancel the zeros to get $\frac{64}{110}$. Dividing both numerator and denominator by $2$, we arrive at our final answer:

$$\mathbf{P = \frac{32}{55}}$$

This journey shows that even the most complex problems become manageable when you break them down into systematic, logical steps. Keep practicing, and you will master this!

Visualized Solution

Initial Configuration

The Transfer Logic

Case 1: Red to Red Setup

Case 1: Calculation

Case 2: Red to Black Setup

Case 2: Calculation

Case 3: Black to Red Setup

Case 3: Calculation

Case 4: Black to Black Setup

Case 4: Calculation

Total Probability Setup

Adding the Numerators

Simplifying the Fraction

Final Answer

The Sigma Insight: Total Probability Theorem

The Anatomy of the Urn Problem

The Four Paths of Destiny

The Synthesis

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