Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the probability distribution of the number of successes, when getting a spade is considered a success.

Hint: In this question it is given that two cards are drawn simultaneously from a well-shuffled deck of 52 cards. We have to find the probability distribution of the number of successes, when getting a spade is considered a success. So to find the solution we have to know that in a deck of 52 cards, there are 13 spades, and let X be the random variable denoting the number of success and success here is getting a spade for an event when two cards are drawn simultaneously. So we have to find the probability distribution for each random variable.
Complete step-by-step solution:
The number of successes is equal to some spades obtained in each draw. So the possible random variables are,
(i) X=0, i.e, selecting 0 spades.
(ii) X=1, i.e, selecting 1 spade.
(iii) X=2, i.e, selecting 2 spades.

(i) So when X=0, i.e, for selecting 0 spades, we removed all 13 spades from the deck and selected out of 39 cards.
$$P\left( X=0\right) =\dfrac{{}^{39}C_{2}}{{}^{52}C_{2}}$$
  =$$\dfrac{\left( \dfrac{39!}{2!\cdot \left( 39-2\right) !} \right) }{\left( \dfrac{52!}{2!\cdot \left( 52-2\right) !} \right) }$$
  =$$\dfrac{\left( \dfrac{39!}{2!\cdot 37!} \right) }{\left( \dfrac{52!}{2!\cdot 50!} \right) }$$
  =$$\dfrac{39!\cdot 2!\cdot 50!}{2!\cdot 37!\cdot 52!}$$
  =$$\dfrac{39\times 38}{52\times 51}$$=$$\dfrac{19}{34}$$.

(ii) now for X=1, i.e, for selecting 1 spade, we need to select 1 out of 13 spades and another one from 39 cards.
$$P\left( X=1\right) =\dfrac{{}^{13}C_{1}\times^{39} C_{1}}{{}^{52}C_{2}}$$
  =$$\dfrac{13\times 39\times 2}{52\times 51}$$
  =$$\dfrac{13}{34}$$

(iii) for X=2, i,e, for selecting 2 spades, we need to select and 2 out of 13 spades,
$$P\left( X=2\right) =\dfrac{{}^{13}C_{2}}{{}^{52}C_{2}}$$
  =$$\dfrac{13\times 12\times 2}{2\times 52\times 51}$$
  =$$\dfrac{2}{34}$$
Therefore, the probability distribution $P_{i}$ for each random variable $X_{i}$,
$$P_{0}=\dfrac{19}{34}$$, for X=0
$$P_{1}=\dfrac{13}{34}$$, for X=1
$$P_{2}=\dfrac{2}{34}$$, for X=2
Note: So to solve this type of question you have to know that probability P=$$\dfrac{n\left( E\right) }{n\left( S\right) }$$ , where n(E)= number of favourable outcome and n(S)= Total number of outcomes. Also $${}^{n}C_{r}$$ defines choosing r number of quantity from n, which can be written as, $${}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}$$.