Question

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} \

Answer 1

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.