Question 1192425
**1. Determine the Average Calls During Lunch Break**

* **Average calls per hour:** 2 calls
* **Average calls during 15 minutes:** 
    * 15 minutes is 1/4 of an hour
    * Average calls = 2 calls/hour * (1/4) hour = 0.5 calls

**2. Use the Poisson Distribution**

* The number of calls received during a specific time interval follows a Poisson distribution when the events (calls) are independent and occur at a constant average rate.

* **Poisson Distribution Formula:**
    * P(X = k) = (λ^k * e^(-λ)) / k! 
        * where:
            * X is the number of calls 
            * λ is the average number of calls during the interval (0.5 in this case)
            * k is the number of calls we're interested in (0 in this case)
            * e is the base of the natural logarithm (approximately 2.71828)

* **Probability of no calls during lunch break (k = 0):**
    * P(X = 0) = (0.5^0 * e^(-0.5)) / 0! 
    * P(X = 0) = (1 * e^(-0.5)) / 1 
    * P(X = 0) = e^(-0.5) 
    * P(X = 0) ≈ 0.6065

**Therefore, the probability that the driver can complete his 15-minute lunch break before receiving a call is approximately 0.6065 or 60.65%.**