+0

# help pls

0
214
13
+275

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

Oct 16, 2022

#1
+118617
+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
+275
0

so 4 even + 2 even + 0 even?

Oct 16, 2022
#3
+118617
+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
+118617
+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
+275
+1

I understand but the computer says its not 5/16

Oct 16, 2022
#6
+2667
0

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

Oct 17, 2022
#8
+1

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

Guest Oct 17, 2022
#7
+275
0

Oct 17, 2022
#11
+2667
0

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
0

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

Guest Oct 17, 2022