Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly three of the dice show a 2? Express your answer as a common fraction.
+118690 Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly three of the dice show a 2? Express your answer as a common fraction.
1,1,2,2,2,* any order
5C2*6 = 60
Edit: I forgot to take into account that the extra number can be 3,4,5,6 so I have to myultiply 60 by 4 to get 240.
60/6^6 *4= 10*4/6^5 = 10*4/7776 = 5*4/3888 = 5/972
That is what I get.
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I will try and explain how I got 240 permutations
It is easier to think of all the diec being different colours. So, for instance, a red 2 is different from a green 2.
The numbers that you need are 1,1,2,2,2,* where star can be 3,4,5, or 6
First how many ways can I order 1,1,2,2,2.
There are 5 positions, how many ways are there that I can choose 2 of them?
That would be 5C2 = 10 ways.
Now I need to choose a number for the wild card. There are 4 to choose from.
Now Now I need to choose a position for the wild number. There are 6 to choose from.
so there are 10*4*6 = 240 favourable permutations.
And the sample space is 6^6 = 46656
So the prob is 240/46656 = 5/972
The computer is listing 240 such permutations, but don't know how it arrived at it!
