If we roll a die 20 times what are the chances that all numbers 1 through 6 come up?
I thought the size of the space was 6^20 and then the size of the event would be
(6!*6^12) however this is obviously wrong.
Thanks for any help
If we roll a die 20 times what are the chances that all numbers 1 through 6 come up?
I thought the size of the space was 6^20 and then the size of the event would be
(6!*6^12) however this is obviously wrong.
Thanks for any help
I think that reply #2 means use inclusion/exclusion.
Lets play the "back-of-the-book" game.
You look up the answer and see $\displaystyle \sum\limits_{k = 0}^5 {\left( { - 1} \right)^k \binom{6}{k}\left( {\frac{{6 - k}}{6}} \right)^{20} } .$
WHAT THE HECK????
Now it is your responsibly the explain why that answer works.
Can you tell us all what is going on there?
Ah ok so If it as all the ways you can choose the 20 with all 6 die thrown then that is the number of surjections from the set of 6 elements to the set of 20 elements which is
$\displaystyle \Sigma^{5}_0 (-1)^k 6Ck (6-k)^{20}$.
We then simply divide by the total space 6^20 and arrive at what you have there.
Sorry quite simple I was just thinking of it in a completely and much less helpful way.
thanks a lot (sorry for so much editing I made loads of typo's)