If six 6-sided fair dice are rolled, what is the probability that three of the dice show prime numbers, and the other three show composite numbers.

Question

anonymous · Answer

(dice are considered distinguishable)

anonymous · Answer

hmmm the chance is $$6$$ *$$\left(\begin{matrix}6 \ 3\end{matrix}\right)$$$$\times 0.5^{3}\times 0.5^{3}$$

anonymous · Answer

Can you please explain your work?

amistre64 · Answer

how many prime numbers on a die?

anonymous · Answer

There are $$6^{4}$$
Which equals to 1296.
different results for rolling 4 six-sided dice. The only way the product of the four upper faces can be prime is if three of the upper faces show a 1 and the remaining upper face shows a prime number less than 6, namely 2, 3, or 5. Here are the possibilities 

1, 1, 1, 2
1, 1, 1, 3
1, 1, 1, 5
1, 1, 2, 1
1, 1, 3, 1
1, 1, 5, 1
1, 2, 1, 1
1, 3, 1, 1
1, 5, 1, 1
2, 1, 1, 1
3, 1, 1, 1
5, 1, 1, 1 

for 12 possible successful outcomes. 
$$\frac{ 12 }{ 1296 } = \frac{ 1 }{ 108 }$$

anonymous · Answer

Uhh, I'm not sure that is the answer to the question, it doesn't involve the product of 4 dice.

anonymous · Answer

Also, I believe there are 4 prime numbers on each die, 1,2,3,5

amistre64 · Answer

1 is not a prime ,,, so lets say 3 out of 6

anonymous · Answer

Whoops, forgot that 1 only has 1 factor

amistre64 · Answer

lets go with 6 choose 3,  1/2^3 in favor and 1/2^3 not in favor

anonymous · Answer

Where do I go from there though?

amistre64 · Answer

id multiply ....

anonymous · Answer

I'm not quite sure what to do.

amistre64 · Answer

6C3 = 20

20 times, 1/2^3 times, 1/2^3 = 20/2^6

anonymous · Answer

Oh, that's it?

amistre64 · Answer

unless im reading the problem incorrectly ,,, then yes

anonymous · Answer

I got... 5/16

amistre64 · Answer

me too :)

anonymous · Answer

It says it's incorrect?

anonymous · Answer

Its under the topic, "Binomial Probability", if that's any kind of hint.

amistre64 · Answer

thats what we did ...

amistre64 · Answer

6 dice, choose 3 successes

6.5.4
----- = 20 outcomes for 3primes and 3 composites
3.2.1

amistre64 · Answer

yes it is lol ... 3 primes and 3not

hmm, might be overlooking the something;  the number of ways a prime can appear

amistre64 · Answer

222  is different from 532

anonymous · Answer

just to butt in for a second 
one is NOT prime

amistre64 · Answer

lol, we already established that ;)

anonymous · Answer

oh 
did you establish also that this is a simple binomial problem?

anonymous · Answer

yeah, looks like you did that too
back to judge judy for me

amistre64 · Answer

theres like 27runs we can make, but they include repeats

222
223
225
232
233
235
252
253
255

322
323
325
332
333
335
352
353
355

522
523
525
532
533
535
552
553
555


weeding out the repeats

222
223
225
233
235
255

333
335
355

555

looks to be it:  10 of them?

anonymous · Answer

Then wouldn't you have to multiply by the number of combinations of the composite numbers?

amistre64 · Answer

maybe,  ive been staring at this screen way to long today ... brains getting fried :)  try the first response;  and see if 6*(our result) works

anonymous · Answer

Dunno, I only have one try left until it is counted wrong :P

amistre64 · Answer

you can afford to miss one right?

anonymous · Answer

Eh, sorta I guess, the way the system works is, you have to answer a certain amount, but if you miss one, you have to answer I think it's something like... 3-7 more?

anonymous · Answer

Also, I don't get what you mean by taking about the repeats, can't they stay?

amistre64 · Answer

maybe they can, im trying to consider a smaller "table"

amistre64 · Answer

rolling 2 dice gets(1p, 1c) gets us 1/2

rolling 4 dice tho ... i wonder if its still gonna be 1/2

mathmate · Answer

There is a problem with the question.  1 is neither a prime nor a composite.

anonymous · Answer

Yeah, I just now noticed the same thing.

amistre64 · Answer

lol, accursed sets

anonymous · Answer

So just try what you first did without 1? xD

amistre64 · Answer

for a binomial to work we would have to have p+q = 1

3*3*3*2*2*2 = 216 ways to get 3 primes and 3 composites

how many permutation can that create?

there are 6*6*6*6*6*6  outcomes total:  46656

anonymous · Answer

I'm not very good at applying permutations, a lot better at combinations, what would it be, (written like xPy)

amistre64 · Answer

i believe the rule would be:$$\frac{6!}{3!3!}$$

amistre64 · Answer

which is 20 :)

anonymous · Answer

20

amistre64 · Answer

20(216)/6^6 = 5/54 ... its woth a shot

anonymous · Answer

Alright

anonymous · Answer

sweet jesus

anonymous · Answer

Its correct

amistre64 · Answer

lol .... yay!!!!!

anonymous · Answer

After an hour xD

anonymous · Answer

Let me check how they did it

anonymous · Answer

They just did 6c3*(1/2)^3*(1/3)^3

amistre64 · Answer

thats what i thought; but im pretty sure that p+q = 1 in a binomial setup

1/2 + 1/3 is not 1

amistre64 · Answer

maybe a 6, 5 sided dice :)

6C3 (3/5)^3 (2/5)^3 ; but thats off too

anonymous · Answer

Thanks