Rohit wants to open his door with
5
5 keys(out of which
1
1 will open the door) and tries the keys independently and at random. If unsuccessful keys are eliminated from further selection, then Find the expected number of trials required to open the door.
Hint: Suppose Rohit gets the first success at
x
t
h
x
th
trial, i.e., he is unable to open the door in the first
(
x
−
1
)
(x−1) trials.
And, P(he gets first success at second trial)=
(
1
−
1
5
)
×
1
4
(1−
5
1
​
)×
4
1
​

To solve this problem, we'll use the concept of **expected value** in probability.

### Problem Recap:
- Rohit has 5 keys, and only 1 key opens the door.
- He tries the keys independently and at random.
- If a key doesn't work, it's eliminated from further selection.
- We need to find the expected number of trials required to open the door.

### Step 1: Understanding the Problem
Let's denote the expected number of trials required to open the door by \( E(X) \).

If Rohit opens the door on the \( x \)-th trial, this means he failed in the first \( x-1 \) trials and succeeded on the \( x \)-th trial.

### Step 2: Probability of Success on the \( x \)-th Trial
For Rohit to succeed on the \( x \)-th trial:
- The probability that he fails in the first \( x-1 \) trials and succeeds on the \( x \)-th trial is:
  \[
  P(\text{Success at } x\text{-th trial}) = \left(\frac{4}{5}\right) \times \left(\frac{3}{4}\right) \times \dots \times \left(\frac{5-x+1}{6-x+1}\right) \times \frac{1}{6-x}
  \]
  
  Specifically, the probability of getting the first success on the \( x \)-th trial is given by:
  \[
  P(\text{Success at } x\text{-th trial}) = \frac{1}{5}
  \]

This simplifies to:
- 1st trial: \( \frac{1}{5} \)
- 2nd trial: \( \frac{4}{5} \times \frac{1}{4} = \frac{1}{5} \)
- 3rd trial: \( \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{1}{5} \)
- 4th trial: \( \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{5} \)
- 5th trial: \( \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} \times \frac{1}{1} = \frac{1}{5} \)

Each key has an equal probability of being the correct one.

### Step 3: Calculate the Expected Value
The expected number of trials \( E(X) \) is given by:
\[
E(X) = 1 \times \frac{1}{5} + 2 \times \frac{1}{5} + 3 \times \frac{1}{5} + 4 \times \frac{1}{5} + 5 \times \frac{1}{5}
\]
This simplifies to:
\[
E(X) = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3
\]

### Final Answer:
The expected number of trials required to open the door is \( \boxed{3} \).

Do you want more details or have any questions?

---

Here are 5 related questions:
1. How does the number of keys affect the expected number of trials?
2. What if Rohit could try the keys multiple times without eliminating them?
3. How would the expected value change if there were more than one correct key?
4. Can you calculate the probability that Rohit opens the door in exactly two trials?
5. How would the problem change if the keys were not selected randomly?

**Tip:** When dealing with expected values, remember to account for all possible outcomes and their associated probabilities.

Explore the expected number of trials required for Rohit to open his door using 5 keys, where only one key works. Learn the concept of expected value in probability and how it applies to this scenario. Suitable for advanced high school students.

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