In a telephone enquiry system, the number of phone calls regarding relevant enquiry follow poisson distribution with an average of five phone calls during 10-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

Options

A. `(6)/(5^e)`
B. `(5)/(6)`
C. `(6)/(55)`
D. `(6)/(e^5)`

Answer

To solve the problem, we need to find the probability that there are at most one phone call during a 10-minute time period, given that the number of phone calls follows a Poisson distribution with an average of 5 calls in that time frame.

### Step-by-Step Solution:

1. **Understand the Poisson Distribution**:
The Poisson distribution is defined by the formula:
\[ P(X = r) = \frac{e^{-m} m^r}{r!} \]
where \( m \) is the average number of occurrences (mean), \( r \) is the number of occurrences, and \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

2. **Identify the Parameters**:
In this problem, the average number of phone calls \( m = 5 \) (given in the question).

3. **Calculate Probability for \( r = 0 \)**:
We first calculate the probability of receiving 0 calls:
\[ P(X = 0) = \frac{e^{-5} \cdot 5^0}{0!} = e^{-5} \cdot 1 = e^{-5} \]

4. **Calculate Probability for \( r = 1 \)**:
Next, we calculate the probability of receiving 1 call:
\[ P(X = 1) = \frac{e^{-5} \cdot 5^1}{1!} = e^{-5} \cdot 5 \]

5. **Combine the Probabilities**:
To find the probability of at most one call, we sum the probabilities for 0 and 1 call:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) = e^{-5} + 5e^{-5} \]
\[ P(X \leq 1) = e^{-5}(1 + 5) = 6e^{-5} \]

6. **Final Result**:
Therefore, the probability that there is at most one phone call during a 10-minute time period is:
\[ P(X \leq 1) = \frac{6}{e^5} \]

### Summary:
The probability that there is at most one phone call during a 10-minute time period is \( \frac{6}{e^5} \).