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Four standard six-sided dice are rolled. What is the probability that the product of the numbers on the top faces of all four dice is 24? Express your answer as a common fraction.

 Jun 21, 2021
 #1
avatar+2115 
+1

1, 1, 4, 6

1, 2, 2, 6

1, 2, 3, 4

2, 2, 2, 3

 

4/6^4 = 1/324

 

=^._.^=

 Jun 22, 2021
 #2
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Since the sample space of 4 dice is 6^4 =1,296, which counts all the "permutations" of the 4 dice, you should therefore count all the permutations whose product==24, since they are all included and counted in the total of 1296 permutations. Therefore, that number of permutations is:

 

1146  1164  1226  1234  1243  1262  1324  1342  1416  1423  1432  1461  1614  1622  1641  2126  2134  2143  2162  2216  2223  2232  2261  2314  2322  2341  2413  2431  2612  2621  3124  3142  3214  3222  3241  3412  3421  4116  4123  4132  4161  4213  4231  4312  4321  4611  6114  6122  6141  6212  6221  6411  Total =  52

 

 

Then the probability is: 52 / 1296 ==13 / 324

 Jun 22, 2021
 #3
avatar+114044 
+1

Using Catmg's combinations that do not appear to be in question

 

1, 1, 4, 6       

1, 2, 2, 6

1, 2, 3, 4

2, 2, 2, 3

 

Assume all the dice are different in colour.  (That will make it much easier)

Also assume that the dice are rolled one at a time.  It is probably easier to act as if order counts.

That means they are all unique and a green 1 is different from a blue 1

 

The total number of combinations is 6^4=1296,   just as catmg found.

 

Each of those 4 combinations above can be formed in 4! ways.  But I do need to allow for the double counting here

 

1, 1, 4, 6       4!/2 = 12

1, 2, 2, 6        4!/2 =12

1, 2, 3, 4           4!=24

2, 2, 2, 3           4

total  52

 

 

 

So we have a prob of    52/1296 =  13/324

 

I am not 100% sure that this is correct .......      But it is the same as guest found so maybe it is ok.

 Jun 23, 2021
edited by Melody  Jun 23, 2021

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