An honest die is rolled 5 times. What is the probability that you will roll NO 6's?
I now know you can use binomial theorem to get the correct answer but I am failing to see why my original method would not have worked...
I found the total # of ways you could roll zero 6's: $5^5$ (5 ways to roll no 6 on the first roll, 5 ways to roll the second roll, etc.) and then divided it by the total number of ways to roll a die 5 times: $6^5$ (6 ways to roll the first roll, 6 ways to roll the second roll, etc.).
I got $0.40$.
Why is that incorrect?
The correct answer is already mentioned in comments, however I would have reached it on a different approach. The odds to score a 6 in one roll are 1/6 therefore the odds to score anything else but a 6 are 5/6. Since the dice rolling is an event independent of previous or next roll, then the odds to score anything but 6 in 5 consecutive events is ${(\frac{5}{6})}^5$