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 $1/100$.
 A: Your answer is correct if you are doing "three white balls among five chosen balls", but is incorrect if the problem is looking for "at least three white balls".Stricktly according to your formula, $P(A | B) = \frac{P(A \cap B)}{P(B)}$
$B=$ at least three white balls among five chosen balls,  $P(B)=\frac{\begin{pmatrix}5\\3\end{pmatrix}\begin{pmatrix}5\\2\end{pmatrix}+\begin{pmatrix}5\\4\end{pmatrix}\begin{pmatrix}5\\1\end{pmatrix}+\begin{pmatrix}5\\5\end{pmatrix}\begin{pmatrix}5\\0\end{pmatrix}}{\begin{pmatrix}10\\5\end{pmatrix}}=\frac{126}{252}$
$P(A \cap B)=\frac{\begin{pmatrix}5\\5\end{pmatrix}\begin{pmatrix}5\\0\end{pmatrix}}{\begin{pmatrix}10\\5\end{pmatrix}}=\frac{1}{252}$
So $P(A | B) =\frac{\frac{1}{252}}{\frac{126}{252}}=\frac{1}{126}$
A: There are $10\choose5$$ = 252$ ways to choose 5 balls from the box.
Half of those cases will be majority white.
We can say this because an equal number of both is not a possibility.
1 case is all of them white.
$\frac1{126}$ 
A: $$\frac{\frac{C\left( \begin{matrix}
   5  \\
   5  \\
\end{matrix} \right)}{C\left( \begin{matrix}
   10  \\
   5  \\
\end{matrix} \right)}}{\frac{C\left( \begin{matrix}
   5  \\
   3  \\
\end{matrix} \right)C\left( \begin{matrix}
   5  \\
   2  \\
\end{matrix} \right)+C\left( \begin{matrix}
   5  \\
   4  \\
\end{matrix} \right)C\left( \begin{matrix}
   5  \\
   1  \\
\end{matrix} \right)+C\left( \begin{matrix}
   5  \\
   5  \\
\end{matrix} \right)C\left( \begin{matrix}
   5  \\
   0  \\
\end{matrix} \right)}{C\left( \begin{matrix}
   10  \\
   5  \\
\end{matrix} \right)}}=\frac{1}{126}$$
A: The probability that all five balls are white is $\frac{5\cdot 4 \cdot 3 \cdot 2 \cdot 1}{10\cdot 9 \cdot 8 \cdot 7 \cdot 6}$. [One way to think of this is $5/10$ chance to choose the first white ball, $4/9$ for the second ball, and so on. Another way is to consider all the balls to be distinct, and count the number of orderings of five balls that are white ($5!$) and divide by the total number of orderings of five balls ($10\cdot 9 \cdots 6$).]
The probability that exactly four are white can be computed similarly. The denominator is still $10 \cdot 9 \cdots 6$, but the numerator is now $5 \cdot 5 \cdot 5!$: five ways to choose a black ball, five ways to choose the white ball that remains in the box, and $5!$ ways to order them.
Similarly, the numerator for the probability of exactly three white balls is $\binom{5}{2} \binom{5}{3} 5!$.
So, the probability of at least three white balls is $\frac{5!}{10\cdot 9 \cdots 6} (1+25+100)=126 \frac{5!}{10\cdot 9 \cdots 6}$.
Using the formula for conditional probability that you wrote, we get $$\frac{\frac{5!}{10\cdot 9 \cdots 6}}{126 \frac{5!}{10\cdot 9 \cdots 6}} = \frac{1}{126}.$$
