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

Guest Jul 18, 2018

#1**0 **

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

Guest Jul 18, 2018

edited by
Guest
Jul 18, 2018

edited by Guest Jul 18, 2018

edited by Guest Jul 18, 2018

#2**+1 **

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.**

Guest Jul 19, 2018

#3**+1 **

**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:**

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

heureka
Jul 19, 2018