A particular 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 overflow floods in a 10 years interval.

Mean expectation = 2; Probability $(X≤3) =0.86$

The correct answer is Option (2) → Mean expectation = 2; Probability $(X≤3) =0.86$

Given the average event of flood overflow in every 10 years is 2.

So, $λ = 2$.

Required probability = $P(X ≤3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$=\frac{2^0e^{-2}}{0}+\frac{2^1e^{-2}}{1}+\frac{2^2e^{-2}}{2}+\frac{2^3e^{-2}}{3}$

$=e^{-2}\left(1+2+2+\frac{4}{3}\right)$

$= 0.135 ×\frac{19}{3}= 0.045 × 19$

$= 0.855 = 0.86$