Answer:
Explanation:
To solve these problems, we can use the Poisson probability formula:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
P(x; λ) is the probability of x events occurring
e is the base of the natural logarithm (approximately 2.71828)
λ is the average rate of events occurring in the given time period
x is the number of events
(a) Probability of exactly 4 orders arriving in 30 minutes:
The average rate of orders is given as 5 orders per hour. To find the average rate of orders in 30 minutes, we divide it by 2 (since 30 minutes is half an hour):
λ = 5 orders/hour / 2 = 2.5 orders/30 minutes
Using the Poisson probability formula:
P(x = 4; λ = 2.5) = (e^(-2.5) * 2.5^4) / 4!
Calculating this:
P(x = 4; λ = 2.5) ≈ (0.082 * 39.0625) / 24
P(x = 4; λ = 2.5) ≈ 3.22265625 / 24
P(x = 4; λ = 2.5) ≈ 0.134
Therefore, the probability that exactly 4 orders will arrive in 30 minutes is approximately 0.134, or 13.4%.
(b) Probability of at least 2 orders arriving in an hour:
To find the probability of at least 2 orders, we need to calculate the probabilities of having 0 and 1 order and subtract it from 1 (since it's the complement).
Using the Poisson probability formula:
P(x = 0; λ = 5) = (e^(-5) * 5^0) / 0! = e^(-5) ≈ 0.0067
P(x = 1; λ = 5) = (e^(-5) * 5^1) / 1! ≈ 0.0337
P(at least 2 orders) = 1 - P(x = 0) - P(x = 1) ≈ 1 - 0.0067 - 0.0337 ≈ 0.9596
Therefore, the probability of at least 2 orders arriving in an hour is approximately 0.9596, or 95.96%.