Question

If the average number of accidents occurring at a particular junction on a highway in a week is 5, then the probability that at most one accident occurs in a particular week is:

• A. \( \frac{25}{e^5} \)
• B. \( \frac{24}{e^5} \)
• C. \( \frac{121}{e^5} \)
• D. \( \frac{6}{e^5} \)

Hint:

For Poisson distributions, use: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] Summing probabilities for cumulative cases ensures correct probability computation.

Answer 1

The Correct Option is C

Since the number of accidents follows a Poisson distribution with mean \( \lambda = 5 \), the probability mass function is: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] To find \( P(X \leq 1) \), we sum the probabilities: \[ P(X = 0) = \frac{5^0 e^{-5}}{0!} = \frac{e^{-5}}{1} = \frac{1}{e^5} \] \[ P(X = 1) = \frac{5^1 e^{-5}}{1!} = \frac{5}{e^5} \] Thus, the probability: \[ P(X \leq 1) = \frac{1}{e^5} + \frac{5}{e^5} = \frac{6}{e^5} \]