Explanation:
if I understand correctly we have only 3 cards to pick from : 4, 5, 6
my answer bases on that understanding.
we don't need any formal calculation, just common sense :
if the first picked card is the 6 card, and we don't put it back, there is no card left greater than 5 for the second pick.
so, this is an impossible event with the automatic probabilty of 0 or 0%.
formally, this looks like this :
a probability is always the ratio
desired cases / totally possible cases.
100% corresponds to 1 (the sure event = the sum of all possible outcomes). so, the % is always 100 times the actual probability.
to pick the 6 out of the possible 3 cards the probability is
1/3 = 0.333333... = 33.3333...%
one desired card out of the possible 3.
now, we don't put that card back.
we are left with 2 cards : 4, 5
the probability to pull a card greater than 5 out of these 2 cards is
0/2 = 0
0 desired cards out of the possible 2.
the probabilty of this combined event of these 2 (independent !) single events is the multiplication of the individual probabilities :
1/3 × 0/2 = 0/6 = 0 = 0×100% = 0%