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Six 6-sided dice are rolled. What is the probability that three of the dice show prime numbers and the rest show composite numbers?

Jul 18, 2018

#1
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I'm not certain about my original thought!!.

Jul 18, 2018
edited by Guest  Jul 18, 2018
edited by Guest  Jul 18, 2018
#2
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Well, I will try the same method that Melody used  a short while ago!:
2 3 5 1 1 1 =120 permutations...........1
2 3 5 1 1 4 =360...................................2
2 3 5 1 1 6 =360...................................3
2 3 5 1 4 4 =360...................................4
2 3 5 1 4 6 =720...................................5
2 3 5 1 6 6 =360...................................6
2 3 5 4 4 4 =120...................................7
2 3 5 4 4 6 =360...................................8
2 3 5 4 6 6 =360...................................9
2 3 5 6 6 6 =120..................................10
[3 x 120] + [6 x 360] + 720 =3,240 / 6^6 =0.069444......etc.
Unless I made a mistake somewhere!!.

Note: Here is a simple way to do this: 6 x 5 x 4 x 3^3 =3,240 / 6^6 =0.069444...etc.

Jul 19, 2018
#3
+22489
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Six 6-sided dice are rolled.
What is the probability that three of the dice show prime numbers and the rest show composite numbers?

Definition "composite number":

A composite number is a positive integer that can be formed by multiplying together two smaller positive integers.
Equivalently, it is a positive integer that has at least one divisor other than 1 and itself.
The composite numbers are exactly the numbers that are not prime and not a unit.

The composite numbers of a dice are: {4,6}

The prime numbers of a dice are: {2,3,5}

The partitions are: {2, 3, 5, 4, 4 ,4}, {2, 3, 5, 4, 4 ,6}, {2, 3, 5, 4, 6 ,6}, {2, 3, 5, 6, 6 ,6}

The permutations of the partitions are:
$$\begin{array}{|c|l|llr|} \hline & \text{partition} & \text{permutation} \\ \hline 1 & {2, 3, 5, 4, 4 ,4} & \dfrac{6!}{1!1!1!3!} &= 4\cdot 5\cdot 6 &= 120 \\ \hline 2 & {2, 3, 5, 4, 4 ,6} & \dfrac{6!}{1!1!1!2!1!} &= 3\cdot 4\cdot 5\cdot 6 &= 360 \\ \hline 3 & {2, 3, 5, 4, 6 ,6} & \dfrac{6!}{1!1!1!1!2!} &= 3\cdot 4\cdot 5\cdot 6 &= 360 \\ \hline 4 & {2, 3, 5, 6, 6 ,6} & \dfrac{6!}{1!1!1!3!} &= 4\cdot 5\cdot 6 &= 120 \\ \hline &&&& \text{sum}=960 \\ \hline \end{array}$$

The probability is:

$$\begin{array}{|rcll|} \hline && \dfrac{120+360+360+120}{6^6} \\\\ &=& \dfrac{2\cdot (120+360) }{6^6} \\\\ &=& \dfrac{960}{6^6} \\\\ &=& 0.02057613169\quad (2.05761316872\ \%)\\ \hline \end{array}$$

Jul 19, 2018
edited by heureka  Jul 19, 2018
edited by heureka  Jul 19, 2018