Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Ask your questions and receive precise answers from experienced professionals across different disciplines. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.