If a random variable X has a Poisson’s distribution such that\[P(X = 1) = P(X = 2)\], it’s mean and variance are
A) 1,1
B) 2,2
C) 2,\[\sqrt 3 \]
D) 2,4

Hint: To solve this question, at first we must have an idea about the probability of the Poisson distribution formula. Then by substituting the value of \[X = 1\]and in the probability of Poisson distribution function equate them to find the value of mean and variance.
The Poisson’s distribution is the discrete probability distribution of the number of events occurring in a given period of time given the average number of times the event occurs over that time of period. For instance, in a binomial distribution if the number of trials gets larger and larger as the probability of success gets smaller and smaller we obtain a Poisson’s distribution.

Complete step by step answer:

If the average number of time that the event occurs is \[\mu \]and X is the actual number of successes, then the Poisson’s probability is given by,
\[P\left( {X,\mu } \right) = \dfrac{{{e^{ - \mu }}{\mu ^X}}}{{X!}}\] …………………………………. (1)
Where \[\mu \] is the mean number of successes in the given time interval or region of space.
Now for\[X = 1\], the probability of Poisson distribution formula is given by
\[P(X = 1,\mu ) = \dfrac{{{e^{ - \mu }}{\mu ^1}}}{{1!}} = {e^{ - \mu }}\mu \] …………………………… (2)
Now for\[X = 2\], the probability of Poisson distribution formula is given by
\[P(X = 2,\mu ) = \dfrac{{{e^{ - \mu }}{\mu ^2}}}{{2!}} = \dfrac{{{e^{ - \mu }}{\mu ^2}}}{2}\] ……………………….. (3)
But given that
\[P(X = 1) = P(X = 2)\] ………………………………. (5)
Now substituting the value of eq. (2) and eq. (3) in eq. (5), we will get,
\[\begin{gathered}
   \Rightarrow {e^{ - \mu }}\mu = \dfrac{{{e^{ - \mu }}{\mu ^2}}}{2} \\
   \Rightarrow \mu = 2 \\
\end{gathered} \]
………………………………. (6)
Here we got the value of the mean.
We know that the mean and variance of the Poisson distribution is equal. Therefore the variance of the Poisson’s distribution is given by
\[{\sigma ^2} = \mu = 2\] ……………………………….. (7)
So, the correct answer is “Option B”.


Note:
In Poisson’s ratio, only one parameter that is mean \[\mu \]is required to determine the probability of any given event.