Question

A bag contains \(19\) red balls and \(19\) black balls. Two balls are chosen at a time repeatedly and discarded if they are of the same color, but if they are different, black ball is discarded and red ball is returned to the bag. The probability that this process will terminate with one red ball is

• A. \(1\)
• B. \(\dfrac{1}{21}\)
• C. \(0\)
• D. \(0.5\)

Hint:

In ball-removal problems, check whether parity is invariant. If the parity of a color remains unchanged throughout the process, it determines the final remaining color.

Answer 1

The Correct Option is A

Step 1: Observe what happens in each operation.
There are two types of balls: \[ 19\text{ red balls and }19\text{ black balls} \] Two balls are selected at a time.

Step 2: Case 1, both balls are red.
If both selected balls are red, then both are discarded.
So the number of red balls decreases by \(2\).
Hence, the parity of the number of red balls does not change.

Step 3: Case 2, both balls are black.
If both selected balls are black, then both are discarded.
So the number of red balls remains unchanged.
Hence, the parity of the number of red balls again does not change.

Step 4: Case 3, one red and one black ball are selected.
If one red and one black ball are selected, then the black ball is discarded and the red ball is returned.
So the number of red balls remains unchanged.
Again, the parity of the number of red balls does not change.

Step 5: Use parity of red balls.
Initially, the number of red balls is \[ 19 \] Since \(19\) is odd, and the parity of the number of red balls never changes, the number of red balls will always remain odd.

Step 6: Determine the final ball.
At the end of the process, only one ball remains.
Since the number of red balls must remain odd, the last remaining ball must be red.

Step 7: Final conclusion.
Therefore, the probability that the process terminates with one red ball is \[ \boxed{1} \]