What is the probability that all three cards are face cards when you do not replace each card before selecting the next card round your answer to the nearest tenth

Define three events A,B,C such that
A = first card is a face card
B = second card is a face card (assuming event A happens)
C = third card is a face card (assuming events A & B happen)
No replacements are made

There are 3 face cards (J, K, Q) per suit. There are 4 suits. In total, there are 3*4 = 12 face cards out of 52 total
P(A) = probability that event A happens
P(A) = 12/52
P(A) = 3/13

After event A happens, we have 12-1 = 11 face cards out of 52-1 = 51 total
P(B) = 11/51

After B happens, we have 11-1 = 10 face cards out of 51-1 = 50 left over
P(C) = 10/50
P(C) = 1/5

Based on how the events A,B,C are set up, we can form this equation
P(A and B and C) = P(A)*P(B)*P(C)
P(A and B and C) = (3/13)*(11/51)*(1/5)
P(A and B and C) = 11/1105
P(A and B and C) = 0.00995475113122

Rounded to the nearest tenth, the answer is simply 0.0 which is a lousy answer in my opinion because it implies that the answer is truly 0 instead of something really small. Though to be fair, the result is very close to 0. A much better approach is to round to the nearest hundredth to get 0.01; though I would ask your teacher for clarification and guidance.

note: a probability of 0 means that the event is impossible to happen, yet it is not impossible to pull out three face cards in a row. Is it unlikely? Yes. Because the result is roughly 0.00995 which is a really small decimal close to 0. But it's not 0 itself. 

What the heck don’t understand