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avatar+251 

Four standard fair -sided dice are rolled. What is the probability that the sum of the numbers rolled is even?

 Oct 16, 2022
 #1
avatar+118132 
+1

 

50% chance of getting an even on one die

 

4 even     1/(2^4)

2even      [1/(2^4) ]* 4!/(2!2!)

0even      1/(2^4)

 

Now add them together.  That's what I think.

 Oct 16, 2022
 #2
avatar+251 
0

so 4 even + 2 even + 0 even?

 Oct 16, 2022
 #3
avatar+118132 
+1

The only ways you can get an even sum is if 

exactly 4 or 2 or 0 dice are even.

 

4 even     1/(2^4) = 1/16

2even      [1/(2^4) ]* 4!/(2!2!)   = 3/16

0even      1/(2^4) = 1/16

 

1+3+1=5

 

So I get a total probability of   5/16

Melody  Oct 16, 2022
 #9
avatar+118132 
+1

Yes my first answer is wrong because of a careless error   4!/(2!2!) = 6   [not 3]

 

so     1+6+1=8   the answer is 8/16 = 1/2

Melody  Oct 17, 2022
 #4
avatar+251 
+1

I understand but the computer says its not 5/16

 Oct 16, 2022
 #6
avatar+2448 
+1

There are just as many ways to roll even as odd, so the answer is 1/2

 Oct 17, 2022
 #8
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+1

I agree with BuilderBoi:  648 / 1296 = 1/2

Guest Oct 17, 2022
 #7
avatar+251 
0

Builderboi? What about the product

 Oct 17, 2022
 #11
avatar+2448 
+1

The product is different (i think) because to get an even product, you need even x even or even x odd; odd only happens with odd x odd. 

BuilderBoi  Oct 17, 2022
 #12
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0

That is true. There are: 1215 even products + 81 odd products ==1,296 total products.

Guest Oct 17, 2022

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