To solve the given problem, let us break it into steps.

Problem Analysis:

The urn contains:

• 2 white balls (W)
• 2 black balls (B)

John's Procedure:

1. He draws two balls simultaneously.
2. If the two balls are of different colors (1 white, 1 black), he stops.
3. If the two balls are of the same color (2 white or 2 black), he returns both to the urn and repeats the process.

We need to calculate the probability that John stops after exactly three draws.

Step 1: Outcomes for Drawing Two Balls

The total ways to draw two balls from the urn are: $\binom{4}{2} = 6$ The possible pairs are:

1. WW (both white)
2. BB (both black)
3. WB (1 white, 1 black) – or equivalently BW (same event).

Probabilities:

• $P(\text{WW}) = \frac{\binom{2}{2} \cdot \binom{2}{0}}{\binom{4}{2}} = \frac{1}{6}$
• $P(\text{BB}) = \frac{\binom{2}{0} \cdot \binom{2}{2}}{\binom{4}{2}} = \frac{1}{6}$
• $P(\text{WB}) = \frac{\binom{2}{1} \cdot \binom{2}{1}}{\binom{4}{2}} = \frac{4}{6} = \frac{2}{3}$

Step 2: Stopping on the 3rd Draw

To stop on exactly the 3rd draw, the first two draws must result in the same color (WW or BB), and the third draw must result in different colors (WB).

Probability of Failing the First Draw:

If the outcome is WW or BB, the probability of drawing same-colored balls in one attempt is: $P(\text{same color}) = P(\text{WW}) + P(\text{BB}) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

Thus, the probability of returning the balls after the first draw is $\frac{1}{3}$.

Probability of Failing the Second Draw:

The same logic applies to the second draw, so the probability of failing on the second draw is also $\frac{1}{3}$.

Probability of Stopping on the Third Draw:

On the third draw, John must draw balls of different colors (WB). The probability of this outcome is: $P(\text{different colors}) = \frac{2}{3}$

Step 3: Combining Probabilities

The probability of stopping on exactly the third draw is: $P(\text{stop on 3rd}) = P(\text{fail first}) \cdot P(\text{fail second}) \cdot P(\text{success on third})$ Substitute the probabilities: $P(\text{stop on 3rd}) = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{27}$

Final Answer:

The probability that John stops after exactly three draws is: $\boxed{\frac{2}{27}}$

Do you want further details, or is this clear? Here are related questions you can explore:

1. What is the probability that John stops on the first draw?
2. What is the probability that John stops on the second draw?
3. How would the problem change if there were 3 white and 3 black balls in the urn?
4. What is the expected number of draws until John stops?
5. How do combinations like $\binom{n}{k}$ influence probability calculations?

Tip: When working with iterative probability, always check for independent and dependent events carefully!