the probability that exactly 4 dice are sixes given that none are fives. My answer key tells me the answer is (10 Choose 4)(1/5)^4 (4/5)^6.
But isn't conditional probability of P(A|B) = P(A and B)/P(B)? Why isn't this answer divided by (5/6)^10? Any help with my intuition? Thanks!
Edit: Sorry, left out some information. It is rolled ten times.

The most straight forward way to interpret that none are fives is simply consider the possible outcomes reduced.  That is, you know each roll was one of $\{1,2,3,4,6\}$.  So knowing this, the probability that exactly four were $6$ is given as
$$ {10 \choose 4} \left( \frac{1}{5} \right) ^4 \left( \frac{4}{5} \right) ^6 $$
If you'd like to use conditional probability, you can write
$$ P(A|B) = \frac{P(AB)}{P(B)} $$
where $P(B) = \left( \frac{5}{6} \right)^{10}$ and $P(AB) = \left( \frac{5}{6} \right)^{10} {10 \choose 4} \left( \frac{1}{5} \right) ^4 \left( \frac{4}{5} \right) ^6 $ which will cancel the factor of $\left( \frac{5}{6} \right)^{10}$ out to give you the answer above :)
