Hey Harold, 
 
Let’s use the following notation:
A – Drawing a spade on the first turn
B – Drawing a spade on the second turn
 
Now, to clarify, the idea is that we draw a card on the second turn, regardless of whether we got a spade initially. We can apply the Multiplication Law to compute the probability of drawing a spade on the second turn and get the correct answer:
1/4 + 3/4*13/51 = 1/4 + 39/204 = (51 + 39)/204 = 90/204 = 15/34
 
Note that this is not really P(B), but rather P(B*), because this is only the case we are interested in, the one where we did not draw a spade on the first go.
 
The law of total probability dictates that P(B) = P(B|A)*P(A) + P(B|A’)*P(A’) = 12/51*1/4 + 13/51*3/4 = 1/4.
 
The idea behind using the Additive Law is a bit more abstract. Since we want to find the probability of getting a spade on either the first or the second turn, we are looking for their union. Thus, P(AuB) = P(A) + P(B) – P(AnB).
 
We already established that P(A) = P(B) = 1/4, so we only need to compute P(AnB). Now, if we get a spade on the first go, we have a 12 in 51 chance of getting a spade on the second turn as well. Thus, P(AnB) = 1/4*12/51. If we plug these numbers into the equation, we get:
P(AuB) = P(A) + P(B) – P(AnB) = 1/4 + 1/4 – 1/4*12/51 = 1/2 – 3/51 = (51 – 6)/102 = 45/102 = 15/34.
 
Hope this helps!
Best,
The 365 Team