A lottery sells `n^(2)` tickets and declares n prizes. If a man purchases n tickets, the probability of his winning exactly one prize is ……

Options

A. `1-(.^(n^(2)-n)C_(n))/(.^(n^(2))C_(n))`
B. `(C_(1).^(n^(2)-n)C_(n-1))/(.^(n^(2))C_(n))`
C. `(.^(n)C_(n).^(n^(2)-n)C_(0))/(.^(n^(2))C_(n))`
D. `1-(1)/(.^(n^(2))C_(n))`

Explanation

To find the probability of winning exactly one prize when a man purchases \( n \) tickets from a total of \( n^2 \) tickets with \( n \) prizes, we can follow these steps:

### Step 1: Determine the total number of tickets and prizes
- Total tickets = \( n^2 \)
- Total prizes = \( n \)

### Step 2: Calculate the number of ways to win exactly one prize
To win exactly one prize, the man must:
1. Win one of the \( n \) prizes.
2. Lose for the remaining \( n - 1 \) tickets.

The number of ways to choose 1 winning ticket from \( n \) prizes is \( \binom{n}{1} = n \).

### Step 3: Calculate the number of ways to choose losing tickets
The man has purchased \( n \) tickets in total. If he wins 1 prize, he must lose on the remaining \( n - 1 \) tickets. The total number of tickets is \( n^2 \), and since he has already won 1 prize, there are \( n^2 - n \) tickets left that are not prizes.

The number of ways to choose \( n - 1 \) losing tickets from the remaining \( n^2 - n \) tickets is given by \( \binom{n^2 - n}{n - 1} \).

### Step 4: Calculate the total number of ways to choose any \( n \) tickets
The total number of ways to choose any \( n \) tickets from \( n^2 \) tickets is \( \binom{n^2}{n} \).

### Step 5: Calculate the probability
The probability \( P \) of winning exactly one prize is given by the ratio of the number of favorable outcomes to the total outcomes:

\[ P = \frac{\text{Number of ways to win exactly one prize}}{\text{Total number of ways to choose } n \text{ tickets}} = \frac{n \cdot \binom{n^2 - n}{n - 1}}{\binom{n^2}{n}} \]

### Step 6: Simplify the expression
Using the property of binomial coefficients, we can simplify the expression:

\[ P = \frac{n \cdot \frac{(n^2 - n)!}{(n - 1)!(n^2 - n - (n - 1))!}}{\frac{(n^2)!}{n!(n^2 - n)!}} \]

This simplifies to:

\[ P = \frac{n \cdot (n^2 - n)! \cdot n!}{(n - 1)! \cdot (n^2)!} \]

### Final Probability Expression
The final expression for the probability of winning exactly one prize is:

\[ P = \frac{n^2 \cdot (n^2 - n)!}{(n - 1)! \cdot (n^2)!} \]