The three cards are drawn simultaneously from the pack of $52$ cards. If one parameter of the card that is ‘aces’ is drawn randomly then find the probability of three ‘aces’ in the pack of$52$.
A. $\dfrac{1}{{5525}}$
B. $5525$
C. $\dfrac{{5525}}{2}$
D. $\dfrac{{5525}}{1}$

Hint:We will use the definition of probability for each iteration/s for a given number of outcomes, that is mathematically indented formulae that to be analyzed with the required outcome or solution. To find the desired outcome/s we will use the concept of combination i.e. $^n{C_r} = \dfrac{{n!}}{{r!(n - r)}}$ and then substituting in the definition of getting the required probability i.e. $\text{Probability} = \dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$ respectively.

Complete step by step answer:
Since, the pack of card is fair, all the outcomes of the pack are equally probable; the probability of ‘aces’ is equally authorized with probability of any other outcomes say, ‘king’,’ queen’, ‘jack’, ‘$10$’, ‘$9$’, ‘$8$’, ‘$7$’, ‘$6$’, ‘$5$’, ‘$4$’, ‘$3$’, ‘$2$’and ‘aces’ respectively.As a result, to find the total number of possible cases of getting ‘aces’ in the pack, according to the combinational statement in the mathematics, we have
$ \Rightarrow {\text{Total outcomes = }}{{\text{ }}^n}{{\text{C}}_r}$
Where, ‘n’ is total outcomes and ‘r’ is required outcome of respective parameter
$ \Rightarrow {\text{Total outcomes = }}{{\text{ }}^{52}}{{\text{C}}_3}$
Substituting the values in$^n{C_r} = \dfrac{{n!}}{{r!(n - r)}}$ the above equation becomes,
$ \Rightarrow {\text{Total outcomes = }}\dfrac{{52!}}{{3!(52 - 3)!}}$
Solving the equation predominantly, we get
$\Rightarrow {\text{Total outcomes = }}\dfrac{{52!}}{{3! \times 49!}} \\
\Rightarrow {\text{Total outcomes = 22100}} \\ $
Since, the above equation is solved by the definition of factorial which seems to be the multiplication of a respective number preceding every number in the sequence of it (till one). But, here we have asked to find the chance of any $3$ outcomes of ‘aces’ in the set of cards out of total $4$ possible outcomes of ‘aces’, Similarly, for finding the probability (can also be treated as favorable outcome), we get
$\Rightarrow {\text{Favorable outcomes = }}{{\text{ }}^4}{{\text{C}}_3} \\
\Rightarrow {\text{Favorable outcomes = }}\dfrac{{n!}}{{r!(n - r)!}} \\ $
Substituting the values in the above equation, we get
\[\Rightarrow {\text{Favorable outcomes = }}\dfrac{{4!}}{{3!(4 - 3)!}} = \dfrac{{4!}}{{3!}} \\
\Rightarrow {\text{Favorable outcomes = 4}} \\ \]
Now, since we know that,
$\text{Probability} = \dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
As a result, substituting both the calculated values in the above formula to get the desired outcome, we get
\[\Rightarrow \text{Probability}= \dfrac{4}{{22100}} \\
\therefore \text{Probability} = \dfrac{1}{{5525}} \\ \]
Hence, the correct option is A.

Note:The probability of an event A is denoted by ‘P(A)’ and the basic formula of finding probability is $P(A) = \dfrac{{n(s)}}{{n(A)}}$, where n(s) denotes the favorable outcomes and n(A) denotes the total number of outcomes for an respective event. Getting rid of misconceptions one must know the basic formulae of solving such application based problems like $^n{C_r} = \dfrac{{n!}}{{r!(n - r)}}$ respectively.