If the number of telephone calls an operator receives between 10:00 am to 10:10 pm follow poisson distribution with mean 3. The probability that the operator receives one call during that interval the next day is

Options

A. `(e^(3))/(3)`
B. `e^(-3)`
C. `(e^(-3))/(3)`
D. `3e^(-3)`

Answer

To solve the problem, we need to find the probability that an operator receives one telephone call between 10:00 am and 10:10 pm, given that the number of calls follows a Poisson distribution with a mean (λ) of 3.

### Step-by-Step Solution:

1. **Identify the Poisson Distribution Formula**:
The probability of a Poisson random variable \( X \) taking a value \( x \) is given by the formula:
\[ P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!} \]
where:
- \( \lambda \) is the mean number of occurrences (in this case, 3).
- \( x \) is the number of occurrences we want to find the probability for (in this case, 1).

2. **Set the Values**:
From the problem, we have:
- \( \lambda = 3 \)
- \( x = 1 \)

3. **Substitute the Values into the Formula**:
Now we substitute \( \lambda \) and \( x \) into the Poisson formula:
\[ P(X = 1) = \frac{3^1 e^{-3}}{1!} \]

4. **Calculate Each Component**:
- Calculate \( 3^1 = 3 \).
- Calculate \( 1! = 1 \).
- The exponential term \( e^{-3} \) remains as is.

5. **Combine the Results**:
Now, substituting these values back into the equation gives:
\[ P(X = 1) = \frac{3 \cdot e^{-3}}{1} = 3 \cdot e^{-3} \]

6. **Final Result**:
Thus, the probability that the operator receives one call during that interval is:
\[ P(X = 1) = 3 e^{-3} \]

### Conclusion:
The correct answer is \( 3 e^{-3} \).