A die is rolled n times. What is the probability of getting at least one 5 and at least one 6

A die is rolled n times. What is the probability of getting at least one 5 and at least one 6

[eqn]1\,-\,\left(\frac23\right)^n[/eqn]
Tell me if you have more homework. :)

You mean
[math]
1-\left( \frac{2}{3} \right)^n
[/math]
which is not correct. It is the probability that there is at least one 5 OR one 6.

Then move your ass and do the shitty sum you get from the law of total probability if you're so smart.

The prob. for exactly f 5's and s 6's is:
[math]
\left( \frac{2}{3} \right)^{n-f-s} \left( \frac{1}{3} \right)^{f+s} \binom{n}{f} \binom{n-f}{s}
[/math]
so the answer is:
[eqn]
\sum_{\substack{2 \leq f+s \leq n \\ 1 \leq s \\ 1 \leq f}}\left( \frac{2}{3} \right)^{n-f-s} \left( \frac{1}{3} \right)^{f+s} \binom{n}{f} \binom{n-f}{s}
[/eqn]

[ n * 1/6 ] ^2

...

WRONG
WRONG
WRONG
WRONG

The only objectively correct answer

>t. MIT statistics professor

Is this the type of shit covered in a normal intro to probability book? This is probably my biggest weakness in math because I learned it early in undergrad when I was a bad student so never really learned it well and then completely forgot what little info I knew.

[math]\begin{align}P( 5 \wedge 6) &= 1-P(\neg(5 \wedge 6)) \\ &=1-P(\neg 5 \vee \neg6) \\ &= 1-(P(\neg 5)+P(\neg 6)-P(\neg 5 \wedge \neg 6)) \\ &= 1-((\frac{5}{6})^n+(\frac{5}{6})^n-(\frac{4}{6})^n) \\ &= \frac{6^n-2*5^n+4^n}{6^n} \end{align}[/math]