Question

The number of accidents per week in a town follows a Poisson distribution with mean 2. If the probability that there are three accidents in two weeks time is \( k e^{-6} \), then the value of \( k \) is:

• A. 18
• B. 36
• C. 27
• D. 9

Hint:

For Poisson distributions, use the formula \( P(k) = \frac{\lambda^k e^{-\lambda}}{k!} \) and adjust for different time intervals or conditions.

Answer 1

The Correct Option is B

Step 1: Understand the Poisson distribution.
The Poisson distribution for the number of events \( k \) occurring in a given interval of time (here, two weeks) is given by: \[ P(k \text{ accidents in two weeks}) = \frac{\lambda^k e^{-\lambda}}{k!}, \] where \( \lambda \) is the average number of events in the interval (here, 2 accidents per week, so \( \lambda = 4 \) for two weeks).

Step 2: Apply the formula for \( k = 3 \).
We are given that the probability is \( k e^{-6} \), so we substitute the values into the Poisson formula: \[ P(3 \text{ accidents}) = \frac{4^3 e^{-4}}{3!} = \frac{64 e^{-4}}{6}. \] Thus, comparing this to \( k e^{-6} \), we find that \( k = 36 \).