# Independence: rolling a dice

• Oct 15th 2013, 09:53 PM
Independence: rolling a dice
If you roll a six-sided die 12 times what is the probability that you get all six numbers at least once?

I feel like the denominator would be 6^12

but then to calculate the probability that each one is selected at least once is what i'm not sure about.
• Oct 16th 2013, 06:58 PM
chiro
Re: Independence: rolling a dice

Hint: You should check out the multi-nomial distribution and calculate the appropriate probability:

Multinomial distribution - Wikipedia, the free encyclopedia
• Oct 16th 2013, 07:42 PM
SlipEternal
Re: Independence: rolling a dice
Figure out the probability that at least one number is not rolled at all. It is the complement of that. So, choose one number out of six. Then, figure out the probability that you do not roll that number for any of the rolls.
Something like this: But it is not quite right...
$\displaystyle 1-\binom{6}{1}\left(\dfrac{5}{6}\right)^{12}$
• Oct 16th 2013, 07:54 PM
SlipEternal
Re: Independence: rolling a dice
Oh, I know what I did wrong. You need inclusion/exclusion for this.
$\displaystyle 1-\binom{6}{1}\left(\dfrac{5}{6}\right)^{12} + \binom{6}{2}\left(\dfrac{4}{6}\right)^{12} - \binom{6}{3}\left(\dfrac{3}{6}\right)^{12} + \binom{6}{4}\left(\dfrac{2}{6}\right)^{12} - \binom{6}{5}\left(\dfrac{1}{6}\right)^{12}$