Question

Suppose that the average number of accidents occurring weekly on a particular stretch of a highway equals 3. The probability that there is at least two accidents in a week is

• A. 1 \(-\) e\(^{-3}\)
• B. 1 \(-\) 2e\(^{-3}\)
• C. 1 \(-\) 3e\(^{-3}\)
• D. 1 \(-\) 4e\(^{-3}\)

Hint:

For Poisson calculations of the form \( P(X \ge n) \):
- Always use the complement \( 1 - \sum_{k=0}^{n-1} P(X = k) \).
- For \( \lambda = 3 \), \( P(X=0) = e^{-3} \) and \( P(X=1) = \lambda e^{-3} = 3e^{-3} \).
This simple pattern lets you quickly sum to \( (1 + \lambda)e^{-\lambda} = 4e^{-3} \).

Answer 1

The Correct Option is D

Step 1: Start from \( P(X=0) \) and build up using the Poisson recursion instead of the formula each time.
For a Poisson variable with mean \( \lambda = 3 \), \( P(X=0) = e^{-\lambda} = e^{-3} \).

Step 2: Step up to \( P(X=1) \) using the recursive relation \( P(X=k+1) = P(X=k)\cdot \frac{\lambda}{k+1} \).
\[ P(X=1) = P(X=0)\cdot\frac{3}{1} = 3e^{-3} \]

Step 3: Add these and take the complement.
\[ P(X \lt 2) = e^{-3} + 3e^{-3} = 4e^{-3} \]
\[ P(X \ge 2) = 1 - 4e^{-3} \]

Step 4: Final Answer.
\[ \boxed{P(X \ge 2) = 1 - 4e^{-3}} \]
Therefore, option (D) is correct.