Probability question?

Question

What's the probability of rolling a die 3 times and getting a 6 every time? What's the probability of rolling a 3 on the forth roll?

What are the key assumptions you are making in this experiment?

Answer 1

P(getting 3 sixes) = 1/6*1/6*1/6 = 1/216

P(getting a 3 on the fourth roll) is the same as the probability of getting a 3 on any roll, 1/6

Now if for some reason, you wanted to know the probability of going 6,6,6,3 for your first 4 rolls, this would be 1/6*1/6*1/6*1/6 = 1/1296

Answer 2

If each number get in a rolling is an "event", I will assume for this answer that each event in your experiment, is independent of others (so one rolling doesnt`affect others). And the probability of getting a number is evenly distributed.
Now, the probability of getting a 6 every time for 3 rollings, is the probability of getting a 6 in the first rolling, a 6 in the second rolling, and a 6 in the third rolling. Because events are independent, this intersections "turns" into a multiplication of probabilities, that`s, what your looking for now is (probability of getting a 6 in the first rollong)*(probability of getting a 6 in the second rollong)*(probability of getting a 6 in the third rolling). Because of the assumptions, each probability I mentioned is 1/6. So the probility of getting a 6 every time is 1/(6^3). The probability of getting a 3 in the forth rollong is the probability of getting a 6 in the three rollings before, and a 3 in the forth. So is the same reasoning, and I hope you can do it for yourself to check if you understood the answer.

Good luck.

Answer 3

The key assumption is that the probability of any one of the numbers coming up on any roll is 1/6.

This is called it being a "fair" die and, more important, that the rolls are "independent" of each other.

Answer 4

P(3 sixes) = 1/6 • 1/6 • 1/6 = 1/216

P(3 on ANY roll) = 1/6

Answer 5

1 in 216.
1 in 6.