Question 1149230
<br>
1+2+3+4+5+6 = 21<br>
P(1) = 1/21
P(2) = 2/21
P(3) = 3/21
P(4) = 4/21
P(5) = 5/21
P(6) = 6/21<br>
You need to get one of each outcome once.  The probability of doing that in any particular order is<br>
{{{(1/21)(2/21)(3/21)(4/21)(5/21)(6/21)}}}<br>
Those six outcomes can occur in any of 6! = 720 different orders.<br>
The probability of getting each outcome once in 6 rolls is<br>
{{{720*((1/21)(2/21)(3/21)(4/21)(5/21)(6/21)) = (720*720)/21^6}}}<br>
That evaluates to 0.0060 to 4 decimal places, or very close to 0.6%.<br>