Conditional Probability - A die is tossed 5 times. A die is tossed 5 times. Given that the die falls six at least once, what's the probability it falls six at least twice?
Let $A$ be the event the die falls six at least once.
$P(A) = 1 - (\frac{5}{6})^{5}$.
Let $B$ be the event the die falls six at least twice.
We want to find $$P(B | A) = \frac{P(B \cap A)}{P(A)}$$
How do I find $P(B\cap A)$?
EDIT:
$$\begin{align} P(B) & = P(A \cap B). \\ & =1 -\left [ \binom{5}{0} \left (\frac{1} {6} \right)^{0}\left (\frac{5}{6} \right)^{5} + \binom{5}{1}\left (\frac{1}{6} \right )^{1} \left (\frac{5}{6} \right)^{4} \right ]. \\  & = 1 - 2\times \bigg(\frac{5}{6}\bigg)^{5}. \\ & = 1 - .803755 \cdots \\ & = 0.196244856 \cdots \end{align}$$ is approx 19%.

Let $X$ be a random variable "number of sixes on 5 throws" then $X \sim Bin(5,1/6)$
Then the question reduces to:
$P(X \geq 2 | X \geq 1) = \frac {P ( X \geq 2 \cap X \geq 1)}{P(X \geq 1)}$
As you said the intersection $X \geq 2 \cap X \geq 1$ is $X \geq 2$
So $P(X \geq 2 | X \geq 1) = \frac {P ( X \geq 2 )}{P(X \geq 1)}$
$P ( X \geq 2 ) = 1 - P(X=1) - P(X=0)$
and
$P ( X \geq 1 ) = 1 - P(X=0)$
$P(X=1)  = 5 \frac{1}{6} (\frac{5}{6})^4 =  (\frac{5}{6})^5$
$P(X=0) = (\frac{5}{6})^5$
and 
$P(X \geq 2 | X \geq 1) = \frac {1-2 (\frac{5}{6})^5}{1-(\frac{5}{6})^5}$
$P(X \geq 2 | X \geq 1) = 0.328101 $
