A random variable $ X $ has Poisson distribution with mean $ 2. $ Then $ P(x > 1.5) $ equals.
(A) $ \dfrac{2} {e^2} \ $
(B) $ 0 $
(C) $ 1 - \dfrac{3} {e^2} \ $
(D) $ \dfrac {3} {e^2} \ $

Hint: Poisson random variable is the number of successes that result from a Poisson experiment. In the Poisson experiment $ P(x,u),x $ takes integral values and $ u $ is mean.

Complete step-by-step answer:
For a Poisson experiment.
 $ P(x,u) = {e^{ - u}}.\dfrac{{{u^x}}}{{x!}} $
Where,
 $ u $ is the average number of successes. Also called as mean.
 $ X $ is the given in the question that means, $ u = 2. $
 $ X $ takes integral values.
Therefore, $ P(x > 1.5) \Rightarrow P(x \geqslant 2) $
We know that, probability of success can also be written as ( $ 1 $ -probability of failure)
As, maximum probability is $ 1. $
 $ \therefore P(x \geqslant 2) = 1 - [P(x = 0) + P(x = 1)] $
 $ = 1 - P(x = 0) - P(x = 1) $
Since, we have
 $ P(x = k) = {e^{ - u}}.\dfrac{{{u^k}}}{{k!}} $
By substituting the values of $ u $ and $ k. $
We get,
$\Rightarrow P(x \geqslant 2) = 1 - {e^{ - 2}}.\dfrac{{{2^0}}}{{0!}} - {e^{ - 2}}.\dfrac{{{2^1}}}{{1!}} $
 $ = 1 - {e^{ - 2}} - {e^{ - 2}} \times 2 $
 $ = 1 - \dfrac{1}{{{e^2}}} - \dfrac{2}{{{e^2}}} $
 $ = 1 - \dfrac{3}{{{e^2}}} $
 $ \therefore P(x > 1.5) = 1 - \dfrac{3}{{{e^2}}} $
Therefore, from the above explanation the correct option is (C) $ 1 - \dfrac{3} {e^2} \ $

Note: To solve this question you need to know that $ x $ takes only integral values. We solved $ P(x \geqslant 2) $ by writing it as $ (1 - P(x = 0) - P(x = 1)) $ because we would have infinite $ x $ for which $ x \geqslant 2. $ Which would not be positive to calculate. Knowing the property of probability that it cannot be more than $ 1 $ helps in such cases.