<s>[INST] Five white balls, five black balls There are 5 white balls and 5 black balls in a box. A visually-challenged man takes 5 balls from the box... What's the probability that all the balls which this individual picked are white if we know for sure that there 3 white balls among those 5 balls he took?
Using the formula $P(A | B) = P(A \cap B)/P(B)$, I get a probability of $1/100$.
 [/INST] The probability that all five balls are white is $\frac{5\cdot 4 \cdot 3 \cdot 2 \cdot 1}{10\cdot 9 \cdot 8 \cdot 7 \cdot 6}$. [One way to think of this is $5/10$ chance to choose the first white ball, $4/9$ for the second ball, and so on. Another way is to consider all the balls to be distinct, and count the number of orderings of five balls that are white ($5!$) and divide by the total number of orderings of five balls ($10\cdot 9 \cdots 6$).]
The probability that exactly four are white can be computed similarly. The denominator is still $10 \cdot 9 \cdots 6$, but the numerator is now $5 \cdot 5 \cdot 5!$: five ways to choose a black ball, five ways to choose the white ball that remains in the box, and $5!$ ways to order them.
Similarly, the numerator for the probability of exactly three white balls is $\binom{5}{2} \binom{5}{3} 5!$.
So, the probability of at least three white balls is $\frac{5!}{10\cdot 9 \cdots 6} (1+25+100)=126 \frac{5!}{10\cdot 9 \cdots 6}$.
Using the formula for conditional probability that you wrote, we get $$\frac{\frac{5!}{10\cdot 9 \cdots 6}}{126 \frac{5!}{10\cdot 9 \cdots 6}} = \frac{1}{126}.$$
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