Question

Vehicular arrival at an isolated intersection follows the Poisson distribution. The mean vehicular arrival rate is 2 vehicles per minute. The probability (rounded off to two decimal places) that at least 2 vehicles will arrive in any given 1-minute interval is \(\underline{\hspace{1cm}}\).

Hint:

For Poisson distributions, the probability of at least \( k \) occurrences is calculated as \( P(X \geq k) = 1 - P(X < k) \).

Answer 1

Correct Answer: 0.58

The probability that at least 2 vehicles will arrive in a 1-minute interval is given by: \[ P(X \geq 2) = 1 - P(X < 2) \] Where \( X \) follows the Poisson distribution with \( \lambda = 2 \) (mean arrival rate). The probability of having fewer than 2 vehicles (i.e., \( X = 0 \) or \( X = 1 \)) is: \[ P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-2} \] \[ P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = 2 e^{-2} \] Thus, the probability that \( X < 2 \) is: \[ P(X < 2) = P(X = 0) + P(X = 1) = e^{-2} + 2 e^{-2} = 3 e^{-2} \approx 3 \times 0.1353 = 0.4059 \] Therefore, the probability that at least 2 vehicles will arrive is: \[ P(X \geq 2) = 1 - 0.4059 = 0.5941 \] Thus, the probability is \( \boxed{0.60} \).