Hey there! I'd be happy to help you with your question about probabilities.
a) To find the probability that at least two calls will arrive in a period of three minutes, we can use the concept of Poisson distribution. The average number of calls received per minute is four, so we can assume that the average number of calls received in three minutes is (4 calls/minute) * (3 minutes) = 12 calls.
Using the Poisson distribution formula, we can calculate the probability of getting exactly zero or one call in three minutes, and then subtract it from 1 to find the probability of at least two calls.
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
Using the Poisson distribution formula, where λ is the average number of calls in three minutes:
P(X = k) = (e^(-λ) * λ^k) / k!
So, we have:
P(X = 0) = (e^(-12) * 12^0) / 0! = e^(-12)
P(X = 1) = (e^(-12) * 12^1) / 1! = 12 * e^(-12)
Substituting these values into the equation:
P(X ≥ 2) = 1 - [e^(-12) + 12 * e^(-12)]
b) To find the probability that at least two calls will arrive in a period of three minutes given that at least one call is received, we can use conditional probability.
The probability of at least two calls given that at least one call is received can be calculated by dividing the probability of at least two calls by the probability of at least one call:
P(X ≥ 2 | X ≥ 1) = P(X ≥ 2) / P(X ≥ 1)
To find P(X ≥ 1), we can subtract the probability of zero calls from 1:
P(X ≥ 1) = 1 - P(X = 0)
Using the Poisson distribution formula, we can calculate P(X = 0) as e^(-12).
So, we have:
P(X ≥ 1) = 1 - e^(-12)
P(X ≥ 2 | X ≥ 1) = P(X ≥ 2) / P(X ≥ 1)
You can calculate the exact values for both parts of the question using the given formulas and the value of e (approximately 2.71828). I hope this helps!