A man has n keys exactly one of which fits the lock. He tries the keys one at a time,

at each trial choosing at random from the keys that were not tried earlier. Find the

probability that the rth key tried is the correct key.

This problem involves finding the probability that the \( r \)-th key tried by the man is the correct one, given that he chooses the keys randomly without replacement.

### Problem Setup:

- There are \( n \) keys, exactly one of which fits the lock.
- The man tries keys one at a time, choosing randomly from the remaining untried keys.
- We want to determine the probability that the \( r \)-th key tried is the correct key.

### Solution:

Since the man is choosing the keys randomly without replacement, we need to calculate the probability that the correct key is selected on the \( r \)-th trial.

1. **Total Number of Keys:**
   - The total number of keys is \( n \), and exactly one key fits the lock.

2. **Probability on the First Attempt:**
   - If the man is on his first attempt, the probability that he selects the correct key is \( \frac{1}{n} \).

3. **Probability on the Second Attempt:**
   - If the correct key is not selected on the first attempt, there are \( n-1 \) remaining keys. The probability that the second key tried is the correct one is \( \frac{1}{n-1} \times \frac{n-1}{n} = \frac{1}{n} \). This result is derived from the fact that the probability of not selecting the correct key on the first attempt is \( \frac{n-1}{n} \), and the probability of selecting the correct key on the second attempt given the first key was incorrect is \( \frac{1}{n-1} \).

4. **General Case (r-th Attempt):**
   - By the same reasoning, the probability that the correct key is selected on the \( r \)-th attempt is:
     - First, the probability that the correct key was not selected in the first \( r-1 \) attempts is \( \frac{n-1}{n} \times \frac{n-2}{n-1} \times \cdots \times \frac{n-(r-1)}{n-(r-2)} \). This simplifies to:
     \[
     \frac{n-r+1}{n}
     \]
     - Second, the probability that the correct key is selected on the \( r \)-th attempt (given that it was not selected in the first \( r-1 \) trials) is \( \frac{1}{n-r+1} \).
  
Thus, the probability that the \( r \)-th key is the correct one is:
\[
P(\text{r-th key is correct}) = \frac{1}{n}
\]

### Conclusion:

The probability that the \( r \)-th key tried is the correct one is the same for every \( r \). Since the trials are random and without replacement, the probability for each trial is uniformly distributed. Therefore, the probability is:
\[
P(\text{r-th key is correct}) = \frac{1}{n}
\]

### Explanation:

This result makes sense because, at every step, the probability that the correct key has not yet been chosen is proportional to the number of remaining keys. Each key has an equal chance of being the correct one, no matter the order of selection.

### Follow-Up Questions:

1. What is the expected number of trials needed to find the correct key?
2. How would the probability change if the keys are selected with replacement?
3. What is the probability that the correct key is found within the first \( r \) trials?
4. How does the scenario change if more than one key fits the lock?
5. What would be the probability distribution of the trial number at which the correct key is found?

### Tip:
In problems involving random selection, uniform probability distributions often arise when the events are symmetric and equally likely, so consider symmetry when analyzing probabilities.

Explore the probability of finding the correct key when choosing randomly from n keys without replacement. Understand the mathematical setup, calculation steps, and conclusions drawn from the analysis. Suitable for Grades 10-12.

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