SOLUTION: Ten is the average number of oil tankers arriving each day at a certain port city. The facilities at the port can handle at most 10 tankers per day. What is the probability that on

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Question 1172793: Ten is the average number of oil tankers arriving each day at a certain port city. The facilities at the port can handle at most 10 tankers per day. What is the probability that on a given day
(a) Tankers have to be turned away?
(b) Tankers have to be facilitated?
(c) No truck arrived at the port?
(d) Exactly 8 trucks have to be facilitated?
Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
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