Probability of drawing any 4 first after that any of clubs Standard 52 cards deck, calculate the probability of drawing any 4 AND after that any of clubs. 
My first intuition is this, there are two possibilities: a) drawing a non clubs 4; or b) drawing a clubs 4, after that you just deduct the card from the deck, so you get:
a) 3/52 * 13/51
b) 1/52 * 12/51
But in the classroom this was resolved as:
a) 4/52 * 13/51
b) 4/52 * 12/51
I don't understand the rationale behind this, is it right/wrong? 
 A: Assuming you mean that two cards are drawn, the probability of drawing both a $4$ and a club is the sum of two cases: (a) the 4 of clubs is drawn, along with any other card; or (b) a club is drawn and a 4 is drawn, neither of which is the 4 of clubs.  The probability of (a) is
$$
p_a = \frac{1}{52} \cdot 2=\frac{1}{26}, 
$$
while the probability of (b) is
$$
p_b = \frac{12}{52} \cdot \frac{3}{51} \cdot 2 = \frac{6}{221}.
$$
Neither of these answers look like either of the answers you gave.

Since the question has changed, I'll add an updated answer.  The question now asks for the probability of drawing first a 4 and then a club.  The probability that the first card is the 4 of the clubs and the second is some other club is
$$
p_a = \frac{1}{52}\cdot\frac{12}{51}.
$$
The probability that the first card is a 4 of some other suit and the second is any club is
$$
p_b = \frac{3}{52}\cdot\frac{13}{51}.
$$
Putting these together gives
$$
p_a + p_b = \frac{12 + 39}{51\cdot 52}=\frac{1}{52}.
$$
A: The probability of drawing first a non-club 4 and then a club is $3/52*13/51$.  The probability of drawing first the 4 of clubs and then a club is $1/52*12/51$.  So the probability of drawing first a 4 and then a club is $3/52*13/51 + 1/52*12/51 = 51/52*51 = 1/52$
I don't understand the second set of equations but maybe it is something like this.
The odds of drawing first a 4 is 4/52.  Then having drawn a four the probability of drawing a club is 13/51 if the 4 wasn't a club (and theres a 3/4 chance of that) or 12/51 if the 4 was a club (and there's a 1/4 chance of that.)
So the probability is $4/52(1/4*12/51 + 3/4*13/52) = 4/52*4*51(12 + 39) = 51/52*51 = 1/52$
