3 cards are drawn at random from a pack of 52 cards. What is the probability that all the three cards are kings?

Hint: Probability is the ratio of possible outcomes to total number of outcomes. Initially there are 52 cards. There are four kings in the pack. So the probability of the card to be king will be the ratio of the number of kings present to total cards. We will simplify the ratio so obtained if required.
Formula used:
\[{\text{Probability = }}\dfrac{{{\text{possible outcomes}}}}{{{\text{total number of outcomes}}}}\]
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step-by-step answer:
Cards are drawn from a pack of well shuffled cards. So the total number of cards are 52. So we can say that total number of outcomes are \[{}^{52}{C_3}\].
 Now there are 4 kings in the pack. But we will draw only 3 cards from the pack so we will have \[{}^4{C_3}\] outcomes.
So the probability can be,
\[{\text{Probability = }}\dfrac{{{}^4{C_3}}}{{{}^{52}{C_3}}}\]
We know that, \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
So we can write,
\[
  {}^{52}{C_3} = \dfrac{{52!}}{{3!\left( {52 - 3} \right)!}} \\
  {}^{52}{C_3} = \dfrac{{52!}}{{3!49!}} \\
  {}^{52}{C_3} = \dfrac{{52 \times 51 \times 50 \times 49!}}{{3!49!}} \\
  {}^{52}{C_3} = \dfrac{{52 \times 51 \times 50}}{6} \\
  {}^{52}{C_3} = 22100 \\
\]
And the numerator can be,
\[
  {}^4{C_3} = \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} \\
  {}^4{C_3} = \dfrac{{4!}}{{3!1!}} \\
  {}^4{C_3} = \dfrac{{4 \times 3!}}{{3!}} \\
  {}^4{C_3} = 4 \\
\]
Thus we can write the probability as,
\[{\text{P}}\left( E \right){\text{ = }}\dfrac{4}{{22100}}\]
On solving we get,
\[{\text{P}}\left( E \right){\text{ = }}\dfrac{1}{{5525}}\]
Probability of drawing the card as king is \[ = \dfrac{1}{{5525}}\]
So, the correct answer is “\[ = \dfrac{1}{{5525}}\]”.

Note: Note that there are four kings in a pack so we can write directly as a ratio of 3 by 52 but that is not the way given. Because there is no specific color mentioned for the king to be drawn. So carefully read the problem always.