What is the probability of the box? Your box of cereal may be a contest winner! It's rattling, which 100% of winning boxes do. Of course 1% of all boxes rattle and only one box in a million is a winner. What is the probability that your box is a winner? 
 A: HINT: Suppose that there are a million boxes. One of them is a winner, and $10~000$ of them rattle. So the probability that a box that rattles is a winner is ... ?
A: Bayes Rule!  : $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$  You want $P(A|B)$ where $A$ is the box being a winner and $B$ is the box rattling.  We know winning boxes rattle and thus $P(B|A)=1$. You have $P(A)=\frac{1}{1000000}$ and $P(B)= \frac{1}{100}$.  Thus: $$P(A|B) = \frac{1*\frac{1}{1000000}}{\frac{1}{100}} = \frac{100}{1000000} = .0001$$
A: On average, out of every 1,000,000 boxes, $\frac{1000000}{100}=$10,000 boxes rattle and $\frac{1000000}{1000000}=$1 box wins.  
Divide the ultimate number by the penultimate number. 
A: This is a simple example for Bayes' theorem:
Denote as A the event when the box is rattling and as B the event when the box is a winner. We then have:
$$p(A) = 0.01$$
and 
$$p(B) = 1 * 10^{-6}.$$
Further, we know that all the boxes that are winners rattle, which writes:
$$p(A|B) = 1,$$
which means the probability of rattling knowing that the box is a winning box.
Now, according to Bayes' theorem we have:
$$p(B|A) = \frac{p(A|B)p(B)}{p(A)} = \frac{p(B)}{p(A)},$$
which gives you the probability of winning, knowing that the box rattles:
$$p(B|A) = 0.0001$$
A: Note that
$$p(A \cap B) = p(B),$$
since every winning box rattles. Therefore according to the definition of conditional probabilities, we have:
$$p(B|A) = \frac{p(A \cap B)}{p(A)} = \frac{p(B)}{p(A)},$$
and we find the same result as when using Bayes' theorem.
A: The correct solution would be $0.0001$ ($1/10000$), wouldn't it? It's late, but it seems to me that Drew Christianson miscalculated and dedocu mixed $p(A)$ and $p(B)$ - correct me please, if I'm wrong.
