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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.

 Feb 6, 2022
 #1
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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

 Feb 6, 2022
edited by Melody  Feb 6, 2022
 #2
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The computer is listing 240 such permutations, but don't know how it arrived at it!

 Feb 6, 2022
edited by Guest  Feb 6, 2022
edited by Guest  Feb 6, 2022
 #3
avatar+117175 
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Yes I just thought of somethin i missed out on.   I get 240 too.

I will edit my first post and talk about it there.

Melody  Feb 6, 2022
 #4
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It can also be computed this way:

 

[6 C 2]  *  [4 C 3]  *  4 (the remaining die can have any of 4 faces: 3, 4, 5, 6] ==240 permutations.

Guest Feb 6, 2022
edited by Guest  Feb 6, 2022

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