1. Key Concepts and Formulas

• Poisson Distribution: This discrete probability distribution models the number of events occurring in a fixed interval of time or space, given a constant average rate of occurrence and independence of events. It is suitable for scenarios like phone calls in a given time, defects per unit area, or arrivals at a service counter.
• Probability Mass Function (PMF): For a Poisson random variable $X$ with an average rate $\lambda$ (lambda), the probability of observing exactly $k$ events in the given interval is: $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ where:
  • $e$ is Euler's number (approximately $2.71828$).
  • $k$ is the specific number of events ($k = 0, 1, 2, \dots$).
  • $k!$ is the factorial of $k$.
• "At Most One" Probability: The phrase "at most one phone call" means that the number of calls can be either $0$ or $1$. Mathematically, this is expressed as $P(X \le 1)$, which is calculated as the sum of probabilities $P(X=0) + P(X=1)$.

2. Step-by-Step Solution

Step 1: Identify the Parameter $\lambda$ (Average Rate) The problem states that the average number of phone calls during a 10-minute time interval is $5$.

• In the context of the Poisson distribution, this average rate is denoted by $\lambda$.
• Therefore, for the specified 10-minute interval, we have $\lambda = 5$.

Step 2: Define the Event and the Probability to be Calculated We are asked to find the probability that there is "at most one phone call" during a 10-minute time period.

• Let $X$ be the random variable representing the number of phone calls in a 10-minute period.
• "At most one phone call" implies that $X$ can be $0$ (no calls) or $1$ (one call).
• So, we need to calculate $P(X \le 1)$.

Step 3: Express $P(X \le 1)$ as a Sum of Individual Probabilities Using the definition of "at most one", we can write: $P(X \le 1) = P(X=0) + P(X=1)$

Step 4: Calculate $P(X=0)$ (Probability of Zero Phone Calls) We use the Poisson PMF with $\lambda = 5$ and $k=0$: $P(X=0) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-5} (5)^0}{0!}$

• Reasoning: Any non-zero number raised to the power of $0$ is $1$ (i.e., $5^0 = 1$). The factorial of $0$ is also $1$ (i.e., $0! = 1$).
• Substituting these values: $P(X=0) = \frac{e^{-5} \cdot 1}{1} = e^{-5}$

Step 5: Calculate $P(X=1)$ (Probability of One Phone Call) Next, we use the Poisson PMF with $\lambda = 5$ and $k=1$: $P(X=1) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-5} (5)^1}{1!}$

• Reasoning: Any number raised to the power of $1$ is itself (i.e., $5^1 = 5$). The factorial of $1$ is $1$ (i.e., $1! = 1$).
• Substituting these values: $P(X=1) = \frac{e^{-5} \cdot 5}{1} = 5e^{-5}$

Step 6: Sum the Probabilities to Find $P(X \le 1)$ Now, we add the probabilities calculated in Step 4 and Step 5: $P(X \le 1) = P(X=0) + P(X=1)$ $P(X \le 1) = e^{-5} + 5e^{-5}$

• Reasoning: We can factor out the common term $e^{-5}$ from both terms. $P(X \le 1) = e^{-5} (1 + 5)$ $P(X \le 1) = 6e^{-5}$
• Reasoning: To match the format of the given options, we can rewrite $e^{-5}$ as $\frac{1}{e^5}$. $P(X \le 1) = \frac{6}{e^5}$

3. Common Mistakes & Tips

• Understanding $\lambda$: Always ensure that the $\lambda$ value used corresponds to the exact interval for which the probability is being calculated. If the interval changes (e.g., from 10 minutes to 20 minutes), $\lambda$ must be adjusted proportionally.
• Factorial of Zero: Remember the convention $0! = 1$. This is a frequent source of error in Poisson distribution calculations.
• Interpreting Keywords: Carefully distinguish between "at most $k$" ($X \le k$), "at least $k$" ($X \ge k$), and "exactly $k$" ($X=k$). Each requires a different approach to summing probabilities.

4. Summary

This problem required us to apply the Poisson distribution to find the probability of "at most one phone call" within a given time interval. We first identified the average rate $\lambda = 5$. Then, we calculated the probabilities of exactly zero calls ($P(X=0) = e^{-5}$) and exactly one call ($P(X=1) = 5e^{-5}$) using the Poisson PMF. Finally, we summed these probabilities to get $P(X \le 1) = 6e^{-5}$, which simplifies to $\frac{6}{e^5}$.

5. Final Answer

The probability that there is at most one phone call during a 10-minute time period is $\frac{6}{e^5}$. This corresponds to option (A), assuming there is a typographical error in the option text and it should read $\frac{6}{e^5}$ instead of $\frac{6}{5^e}$.

The final answer is $\boxed{\text{A}}$.