A bag contains `n` white and `n` red balls. Pairs of balls are drawn without replacement until the bag is empty. Show that the probability that each pair consists of one white and one red ball is `(2^n)/(""^(2n)C_(n))`

Options

A. `1//^(2n)C_(n)`
B. `2n//^(2n)C_(n)`
C. `2n//n!`
D. `2n//(2n!)`

Explanation

To solve the problem, we need to find the probability that each pair drawn from a bag containing `n` white balls and `n` red balls consists of one white and one red ball. We will do this step by step.

### Step-by-Step Solution:

1. **Understanding the total number of ways to draw pairs:**
The total number of balls in the bag is `2n` (n white + n red). We will draw pairs of balls until the bag is empty. The total number of ways to choose 2 balls from `2n` balls is given by the combination formula:
\[ \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1) \]

2. **Calculating the number of favorable outcomes:**
We want to find the number of ways to draw pairs such that each pair consists of one white and one red ball. For the first pair, we can choose 1 white ball from `n` white balls and 1 red ball from `n` red balls. The number of ways to choose the first pair is:
\[ n \times n = n^2 \]
After drawing the first pair, we have `n-1` white balls and `n-1` red balls left. The number of ways to choose the second pair is:
\[ (n-1) \times (n-1) = (n-1)^2 \]
Continuing this process, the number of ways to choose the `k`-th pair is:
\[ (n-k+1)(n-k+1) \]
Therefore, the total number of favorable outcomes when drawing all pairs is:
\[ n^2 \times (n-1)^2 \times (n-2)^2 \times \ldots \times 1^2 = n^2 \times (n-1)^2 \times (n-2)^2 \times \ldots \times 1^2 = (n!)^2 \]

3. **Calculating the total number of ways to draw all pairs:**
The total number of ways to draw all pairs from `2n` balls (without any restriction) is given by:
\[ \frac{(2n)!}{2^n} \]
This is because we are pairing up `2n` balls into `n` pairs, and each pair can be arranged in `2` ways (hence the division by `2^n`).

4. **Finding the probability:**
The probability that each pair consists of one white and one red ball is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\[ P = \frac{(n!)^2}{\frac{(2n)!}{2^n}} = \frac{(n!)^2 \cdot 2^n}{(2n)!} \]

5. **Final expression:**
We can express this probability as:
\[ P = \frac{2^n}{\binom{2n}{n}} \]
where \(\binom{2n}{n}\) is the number of ways to choose `n` positions for the white balls out of `2n` total positions.

### Conclusion:
Thus, we have shown that the probability that each pair consists of one white and one red ball is:
\[ P = \frac{2^n}{\binom{2n}{n}} \]