Correct Answer - Option 1 : 4e

^{-2}
**Formula**

**Poisson distribution = P(X = x) = [ e**^{-m} m^{x}/x! for x = 0, 1, 2, 3 -----]

** [ 0 ; otherwise]**

**Calculation**

Poisson distribution has 2 modes at x = 1 and x = 2 it means they have same frequency at x = 1 and 2and have same probability

P(x = 1) = λ^{1}. e^{-λ}/1!

⇒ P(x = 2) = λ^{2}. e^{-λ}/2!

According to question P(x = 1) = P(x = 2)

λ^{1}. e^{-λ}/1! = λ^{1}. e^{-λ}/2!

⇒ λ^{1}. e^{-λ} = λ^{2}. e^{-λ}/2

⇒ λ = 2

⇒ P(x = 1) = 2^{1}. e^{-2}/1!

⇒ 2e^{-2}

⇒ P(x = 2) = 2^{2}. e^{-2}/2!

⇒ 2.e^{-2}

⇒ P(x) = P(x = 1) + P(x = 2)

⇒ 2e^{-2} + 2.e^{-2}

**∴**** P(x) = 4e**^{-2}