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.
Hints:
Combinatorial way 1: There are $\binom{52}{4}$ equally likely hands of $4$. There are $\binom{39}{4}$ no club hands.
Combinatorial way 2: Imagine dealing the cards one at a time. There are $(52)(51)(50)(49)$ equally likely strings of four distinct cards. How many strings of $4$ cards have no club?
Conditional probability way: Imagine that the cards are dealt one at a time. The probability the first card is not a club is $\frac{39}{52}$.
Given that the first card is not a club, the probability the second card is not a club is $\frac{38}{51}$. So the probability that neither of the first two cards is a club is $\frac{39}{52}\cdot \frac{38}{51}$.
Given that neither of the first two card is a club, the probability the third card is not a club is $\frac{37}{50}$. Continue.
Best Answer
b & c are different scenarios, but you get the same answer, still. b states that you need to find the probability of 'drawing a seven, returning it to the deck and right after pulling a red suit card?'. That means you need to find the probability of drawing a 7 from 52 cards, which is $\frac4{52}$. Now, you need to put the card you drew back into the pile, leaving it with 52 cards again. Now find the probability of drawing a red card. There are 2 colors, red and black, each with equal probability, so this is equal to $\frac12$. Now multiply the probabilities you got. $\frac4{52}$ * $\frac12$ = $\frac1{26}$. So that is the answer to b.
c is very similar, but after you draw the seven, you don't put the 7 back in the pile, so there are only 51 cards left. Now we split this up into 2 cases. If the 7 is red, there is a $\frac2{52}$ chance to draw it. Since we drew a red seven, there are 25 reds to draw out of the remaining 51 cards, so the chance of drawing a red card after drawing a red 7 is $\frac{25}{51}$. Now multiply the probabilities $\frac2{52}$ * $\frac{25}{51}$ = $\frac{25}{1326}$. The second case is that the 7 we drew was black, a chance of $\frac2{52}$. This leaves 26 red cards out of 51 left to draw, so the chance of drawing a red is now $\frac{26}{51}$. Multiply the probabilities to get $\frac2{52}$ * $\frac{26}{51}$ = $\frac1{51}$. Now add the probabilities of both cases to get $\frac{25}{1326}$ + $\frac1{51}$ = $\frac{1}{26}$.