Step-by-step explanation:
this is a kind of trick question, actually.
with whatever we draw, we produce X values as power of 3.
to be precise, we can have only
3⁰ = 1
3¹ = 3
3² = 9
3⁴ = 81
due to the possible combinations of drawn numbers (e.g. 3 cannot be created by a multiplication of 0s, 1s and 2s).
so, mostly, these results cannot be exact factors of 1024.
1024 cannot be divided by 3, nor by 9 nor by 81.
but 1024 is a multiple of 1 (as is every number).
so, we are looking at the probability to get 0 as multiplication result of the numbers on the 2 drawn balls.
the only possibilities are
1 and 0
2 and 0
out of in total 6 (2×3) different outcomes
1 and 0
1 and 1
1 and 2
2 and 0
2 and 1
2 and 2
the probability of this "0" event is again
number of desired outcomes / number of possible outcomes = 2/6 = 1/3
