Question

At a traffic intersection, cars and buses arrive randomly according to independent Poisson processes at an average rate of 4 vehicles per hour and 2 vehicles per hour, respectively. The probability of observing at least 2 vehicles in 30 minutes is \(\underline{\hspace{2cm}}\) . (round off to two decimal places)

Hint:

The Poisson distribution can be used to calculate the probability of observing a given number of events in a fixed interval of time.

Answer 1

Correct Answer: 0.78 - 0.82

Let the total rate of arrival be the sum of the rates for cars and buses:
\[ \lambda = 4 + 2 = 6 \, \text{vehicles per hour}. \] For 30 minutes, the rate becomes:
\[ \lambda_{30} = 6 \times 0.5 = 3 \, \text{vehicles}. \] The probability of observing at least 2 vehicles is given by the Poisson distribution:
\[ P(X \geq 2) = 1 - P(X < 2) = 1 - \left[ P(X = 0) + P(X = 1) \right]. \] Using the Poisson probability formula:
\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}. \] For \( \lambda = 3 \), we calculate:
\[ P(X = 0) = \frac{3^0 e^{-3}}{0!} = e^{-3} \approx 0.0498, \] \[ P(X = 1) = \frac{3^1 e^{-3}}{1!} = 3e^{-3} \approx 0.1494. \] Thus:
\[ P(X \geq 2) = 1 - (0.0498 + 0.1494) = 1 - 0.1992 = 0.8008. \] Thus, the probability is \( \boxed{0.81} \).