I got 5/108.
(1/6)² * (5/6)¹ * 3!/(2!1!) * [(1/3)² * (2/3)¹ * 3!/(2!1!)] * 3
What am I missing here?
Why are you considering permutations for the “either 1 or 2 appears in each of the remaining three rolls of the die”.
Edit: Sorry i edited my “new answer” into this comment for some weird reason
What? i am considering it a combination. ( 1 or 2 i.e. two possibilities (2/6)^2* (4/6)^1 *3!/2!1!)*3 since it can occur three times.
I’m saying that you’re treating this like a binomial when it isn’t. You permute stuffs in binomial and just because you wrote it with “combination” doesn’t mean it’s not a permutation.
Let (D_1, D_2, D_3) represent the results of the three die. You’re looking for instances where D_1, D_2, D_3 are all either 1 or 2. There are 6^3 possible tuples in total in which 2^3 are of interest, so it’s just \frac {1}{27} for the last three rolls.
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Thanks for that. Now I have to do the probability all over again. I have no idea what you are saying (it’s my weakness).
Okay basically put, the “concept” for the first three rolls is different from the “concept” for the last three rolls. Your formula is relevant for the first three rolls but not the next 3.
If you want, you can just sidestep using the “formula” altogether and just think like this:
For the first three rolls, you consider all arrangements where 5 appears exactly two times, so like:
(5, 5, S), (5, S, 5), (S,5,5) → You care about these cases where S is any number other than 5, right? This is because you want exactly 2 5’s so having S = 5 goes against what we want. So in total, you’re looking at 5 * 3 = 15 cases.
In total, there are 6^3 cases (hopefully u see this too?) but you only care about those 15 cases above 15/6^3 is the required probability for the first three rolls.
If you get all of this then try to repeat this same argument for the next three rolls and then yeah you should have your answer.
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Thanks, i got it now.