A student has to match three historical events Dandi March, Quit India movement and Mahatma Gandhi Assassination with the years 1948,1930 and 1942. The student has no knowledge of the correct answers and decided to match the events and years randomly. If x denotes the number of correct answers he gets, then mean of x is

Options

A. `0.5`
B. 1
C. `1.5`
D. 2

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the events and years
We have three historical events:
1. Dandi March
2. Quit India Movement
3. Mahatma Gandhi Assassination

And the corresponding years:
1. 1948
2. 1930
3. 1942

### Step 2: Determine the total number of ways to match events with years
The total number of ways to match the three events with the three years is calculated using the factorial of the number of events (or years). Since there are 3 events, the total number of arrangements is:
\[ 3! = 3 \times 2 \times 1 = 6 \]

### Step 3: Calculate the probabilities for different values of x
- **Probability of x = 3 (all correct)**: There is only 1 way to match all correctly (the correct arrangement), so:
\[ P(x = 3) = \frac{1}{6} \]

- **Probability of x = 2 (two correct)**: If two are correct, the third must also be correct (since there are only three events), so:
\[ P(x = 2) = 0 \]

- **Probability of x = 1 (one correct)**: The student can match one correctly in 3 different ways (choosing any one of the three events to match correctly), so:
\[ P(x = 1) = \frac{3}{6} = \frac{1}{2} \]

- **Probability of x = 0 (none correct)**: The remaining arrangements where none are correct can be calculated as:
\[ P(x = 0) = \frac{2}{6} = \frac{1}{3} \]

### Step 4: Calculate the mean (expected value) of x
The mean (or expected value) of x is calculated using the formula:
\[ E(x) = \sum (x \cdot P(x)) \]
Substituting the values we calculated:
\[ E(x) = 0 \cdot \frac{2}{6} + 1 \cdot \frac{3}{6} + 2 \cdot 0 + 3 \cdot \frac{1}{6} \]
Calculating this gives:
\[ E(x) = 0 + \frac{3}{6} + 0 + \frac{3}{6} = \frac{6}{6} = 1 \]

### Final Answer
The mean of x (the expected number of correct matches) is:
\[ \boxed{1} \]