<s>[INST] 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.
 [/INST] 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 :)
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