# 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?

-

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.

-

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 ... ?

-
I was going to post something like this myself. A good way to approach these problems is to just make up a population size—in this case 1,000,000—and then compute the sizes of the sub-populations. This concrete approach is often easier to reason about than the corresponding (and mathematically equivalent) probabilities. –  MJD Sep 24 '12 at 23:49

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: $$P(A|B) = \frac{1*\frac{1}{1000000}}{\frac{1}{100}} = \frac{100}{1000000} = .0001$$

-
It's a bit simpler to just use the definition of conditional probability: $P(A|B)={P(A\cap B)\over P(B)}$ and to note that $P(A\cap B)=P(A)$, here. –  David Mitra Sep 24 '12 at 21:38
Definitely, but I think it's also useful to do some easy problems with more complicated machinery, just to see how things work. –  Drew Christianson Sep 24 '12 at 21:40

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: $$p(B|A) = 0.0001$$

-
Corrected after suggestion from Christian...there was a $0$ to much in the result...sorry for that! –  jojo Sep 26 '12 at 20:51

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.

-
YES you are right –  Rohit Sep 25 '12 at 6:20
Well I tried to edit the corresponding post from dedocu, but I 'changed to much' and my edit was rejected, so the wrong solution will stay there. At least I could correct the other entry. –  Christian Sep 25 '12 at 11:24

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.

-