I have a feeling your teacher/professor intended for you to learn the following (we first introduce some notation):
For two sets $A, B$ in a common universe $U$, define their union as
$$
A \cup B = \{x \in U : x \in A \text { or } x \in B\}.
$$
Define their intersection as:
$$
A \cap B = \{x \in U : x \in A \text{ and } x \in B\}.
$$
It is important not to get too wrapped in the English here. Being in $A \cup B$ simply means being in one of the two sets (or possibly both). Being in $A \cap B$ simply means being in set $A$ and being in set $B$ at the same time.
Finally, a finite set $A$ with $k$ elements ($k$ things in the set) has cardinality $k$, and this is written as $|A| = k$, or $\#A = k$ or even sometimes, $n(A) = k$.
Therefore:
$$
\{1,3,5\} \cup \{1,2,3\} = \{1,2,3,5\},
$$
while
$$
\{1 ,3 , 5 \} \cap \{ 1 , 2, 3 \} = \{ 1, 3\}.
$$
Also,
$$
| \{ 1,3,5\}| = 3,
$$
while
$$
| \{ 1, 3\} | = 2.
$$
Now, what your teacher probably wanted you to learn was the following "rule":
$$
|A \cup B| = |A| + |B| - |A \cap B|.
$$
This is easy to see it is true, since to count the number of elements that are in either $A$ or $B$ (or possibly both $A$ and $B$), you count the number of elements in $A$, add to it the number of elements in $B$, and then subtract the stuff you double counted, which is precisely the elements in $A \cap B$.
Therefore, if you want to find the number of sides of a die that are (say) even or prime, you count the number of sides which are even (there are $3$ such sides - namely $2$ , $4$, $6$) add to it the number of sides which are prime (again, there are $3$ such sides - namely $2$, $3$, $5$), and then subtract the sides which we double counted (we counted the side with the number $2$ twice).
Therefore, there are $3 + 3 - 1 = 5$ sides of a die which are even or prime.
Now, you can take this strategy and count the number of cards in a deck which are either non face cards or clubs.
Best Answer
Looks to me like you're overcomplicating this. Answering the following question should help you
The answer to this question is no. It doesn't matter, because every card has the same properties relative to the others (3 cards with same value, 12 with same suit, 25 with same colour, etc).
So assume your card is the Ace of Spades, for argument's sake.
Then a "winning" card is any of the following:
So there are $39$ winning cards out of $51$, giving a probability of $\frac{39}{51}$