16. Online booking system of food channel receives an average of two orders in every four minutes. Assuming an approximate Poisson distribution that three or more orders will be received during a period of seven minutes? A) 0.3712 B) 0.4679 C) 0.5321 D) 0.3208

To solve this problem, we first need to calculate the expected number of orders in seven minutes. Given that the average rate is two orders every four minutes, we can deduce that in seven minutes, the average number of orders \( \lambda \) would be: \[ \lambda = \frac{2 \text{ orders}}{4 \text{ minutes}} \times 7 \text{ minutes} = 3.5 \text{ orders} \] Next, we use the Poisson probability formula to calculate the probability of receiving three or fewer orders: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] We will calculate this for \(k = 0, 1, 2, \) and \(3\) and sum these probabilities to find \(P(X \leq 3)\). After that, we can find \(P(X \geq 3) = 1 - P(X \leq 2)\). Performing these calculations, you should reach the solution: The correct answer is **C) 0.5321.**