Question 1172793
Let's solve this problem using the Poisson distribution.

**Given:**

* Average number of tankers arriving per day (λ) = 10
* Port capacity = 10 tankers per day

**a. Probability Tankers Have to Be Turned Away**

* Tankers are turned away if more than 10 arrive.
* We need to find P(X > 10), where X is the number of tankers arriving.
* P(X > 10) = 1 - P(X ≤ 10)
* We can use the Poisson distribution formula: P(X = k) = (e^(-λ) * λ^k) / k!
* P(X ≤ 10) = Σ [ (e^(-10) * 10^k) / k! ] for k = 0 to 10
* Using a calculator or statistical software:
    * P(X ≤ 10) ≈ 0.5830
* P(X > 10) = 1 - 0.5830 ≈ 0.4170

**b. Probability Tankers Have to Be Facilitated**

* Tankers are facilitated if 10 or fewer arrive.
* This is P(X ≤ 10).
* From part (a), we already found P(X ≤ 10) ≈ 0.5830

**c. Probability No Truck Arrived at the Port**

* We need to find P(X = 0).
* P(X = 0) = (e^(-10) * 10^0) / 0!
* P(X = 0) = e^(-10) * 1 / 1
* P(X = 0) ≈ 0.0000454

**d. Probability Exactly 8 Trucks Have to Be Facilitated**

* We need to find P(X = 8).
* P(X = 8) = (e^(-10) * 10^8) / 8!
* P(X = 8) = (e^(-10) * 100000000) / 40320
* P(X = 8) ≈ 0.1126

**Summary:**

* **(a) Probability tankers have to be turned away:** ≈ 0.4170
* **(b) Probability tankers have to be facilitated:** ≈ 0.5830
* **(c) Probability no truck arrived at the port:** ≈ 0.0000454
* **(d) Probability exactly 8 trucks have to be facilitated:** ≈ 0.1126