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 $e^{-2} = 0.13534$)

Mean = 2; Probability = 0.86

The correct answer is Option (1) → Mean = 2; Probability = 0.86

Here $λ = 2$ (given)

So, mean expectation = $λ = 2$

$P(X = r) =\frac{e^{-λ}λ^r}{r}=\frac{e^{-2}(2)^r}{r}$

Now, $P(X ≤3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$=e^{-2}\left[1+2+2+\frac{4}{3}\right]=e^{-2}×\frac{19}{3}$

$=0.13534×\frac{19}{3}=0.8571=0.86$