Question

A river near a small town floods and overflows twice in every 10 years on an average. Assuming that the Poisson distribution is appropriate, what is the mean expectation? Also, calculate the probability of 3 or less overflows and floods in a 10-year interval.
[Given \( e^{-2} = 0.13534 \)]

Hint:

For Poisson distribution problems, use the formula \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \). To calculate \( P(X \leq k) \), sum the probabilities from \( P(X = 0) \) to \( P(X = k) \).

Answer 1

Step 1: The Poisson distribution formula is: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \] where \( \lambda \) is the average rate of occurrence, \( k \) is the number of occurrences, and \( e \) is the base of the natural logarithm. 
Step 2: Mean expectation (\( \lambda \)): The average number of floods in 10 years is \( \lambda = 2 \). 
Step 3: Probability of 3 or fewer overflows (\( P(X \leq 3) \)): \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). \] Using the Poisson formula: - For \( P(X = 0) \): \[ P(X = 0) = \frac{2^0 e^{-2}}{0!} = \frac{1 \cdot 0.13534}{1} = 0.13534. \] - For \( P(X = 1) \): \[ P(X = 1) = \frac{2^1 e^{-2}}{1!} = \frac{2 \cdot 0.13534}{1} = 0.27068. \] - For \( P(X = 2) \): \[ P(X = 2) = \frac{2^2 e^{-2}}{2!} = \frac{4 \cdot 0.13534}{2} = 0.27068. \] - For \( P(X = 3) \): \[ P(X = 3) = \frac{2^3 e^{-2}}{3!} = \frac{8 \cdot 0.13534}{6} = 0.18045. \] 
Step 4: Add the probabilities: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). \] Substitute the values: \[ P(X \leq 3) = 0.13534 + 0.27068 + 0.27068 + 0.18045 = 0.85715. \] 
Final Answers: - Mean expectation: \( \lambda = 2 \). - Probability of 3 or fewer overflows: \( P(X \leq 3) = 0.85715 \) or approximately \( 85.72\% \).