Question 191368: What is the probability of rolling a die three times, and getting outcomes of 1, 2, and 3, in this order.
I tried to work as hard as I can on this problem but I can't find a way to know the probability that will be in THE SAME ORDER as the question asks.
I did (1/6) x (1/6) x (1/6) and that got me 1/216
I don't know... Should I do (1/6)^3 x (1/6)^3 x (1/6)^3?
I hope this doesn't count as one of my daily problem limit because I really tried my hardest.
Thanks
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You were right the first time. You have three independent events, each with a probability of  , hence:
\left({1 \over 6}\right)\left({1 \over 6}\right) = {1 \over 216})
If the order didn't matter, then you would have:
A 1 in 2 chance (i.e. 3 out of 6) ways to get a 1, a 2, or a 3, then
A 1 in 3 chance (i.e. 2 out of 6) ways to get one of the two remaining numbers, then
A 1 in 6 chance of getting the last remaining number, for:
\left({1 \over 3}\right)\left({1 \over 6}\right) = {1 \over 36})
Now this all makes sense if you consider that the probability of getting 1, 2, and 3 in any other specific order is also  , and there are 6 different orders that you can have for three numbers (3! = 6), and then:
The exact same probability as if order didn't matter.
John
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