If the variance of the Poisson distribution is 2, find the probabilities for r = 1, 2, 3, 4 and 5 from the recurrence relation of the Poisson distribution. (Use $e^{-2} = 0.1353$)

$P(X=1)=0.2706, P(X=2)=0.2706, P(X=3)=0.1804, P(X=4)=0.0902, P(X=5)=0.0361$

The correct answer is Option (1) → $P(X=1)=0.2706, P(X=2)=0.2706, P(X=3)=0.1804, P(X=4)=0.0902, P(X=5)=0.0361$

As the variance is $λ$, we get $λ = 2$.

$P(0)=e^{-λ}=e^{-2}=0.1353$.

Now we use the recurrence relation of the Poisson distribution

$P(r + 1) =\frac{λ}{r+1}= P(r)$

$P(1) =\frac{λ}{1} P(0)=2(0.1353) = 0.2706$

$P(2) =\frac{λ}{2} P(1) = P(1) = 0.2706$

$P(3) =\frac{λ}{3} P(2) =\frac{2}{3} (0.2706) = 0.1804$

$P(4) =\frac{λ}{4} P(3) =\frac{1}{2} (0.1804) = 0.0902$

$P(5) =\frac{λ}{5} P(4)=\frac{2}{5}(0.0902) = 0.0361$