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: Understanding the Question:
This question asks for the probability of at least two weekly accidents occurring on a highway stretch, given that the average rate of occurrence is 3 accidents per week.
This is a classic Poisson distribution problem because we are modeling the number of rare discrete events occurring within a fixed time interval.

Step 2: Key Formula or Approach: For a Poisson random variable \( X \) representing the number of events, the probability of observing exactly \( k \) events is given by: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where:
\( \lambda \) is the average rate of events per interval (here, \( \lambda = 3 \)).
We need to calculate \( P(X \ge 2) \).

Step 3: Detailed Explanation:

• Use Complementary Probability:
- Finding the probability of "at least two" directly involves an infinite sum: \( P(X=2) + P(X=3) + \dots \).
- Instead, use the complement rule:
\[ P(X \ge 2) = 1 - P(X \lt 2) \]
- The events for \( X \lt 2 \) are \( X = 0 \) and \( X = 1 \):
\[ P(X \lt 2) = P(X = 0) + P(X = 1) \]

• Calculate Individual Probabilities:
- For \( k = 0 \):
\[ P(X = 0) = \frac{e^{-3} 3^0}{0!} = \frac{e^{-3} \cdot 1}{1} = e^{-3} \]
- For \( k = 1 \):
\[ P(X = 1) = \frac{e^{-3} 3^1}{1!} = \frac{e^{-3} \cdot 3}{1} = 3e^{-3} \]

• Sum and Complement:
- Add these values:
\[ P(X \lt 2) = e^{-3} + 3e^{-3} = 4e^{-3} \]
- Compute the final probability:
\[ P(X \ge 2) = 1 - 4e^{-3} \]


Step 4: Final Answer:
The probability of at least two accidents in a week is \( 1 - 4e^{-3} \).
Therefore, the correct choice is option (D).