Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

6 Answers 6

up vote 2 down vote accepted

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.

share|improve this answer

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 ... ?

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

share|improve this answer
    
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: $$p(B|A) = 0.0001$$

share|improve this answer
    
Corrected after suggestion from Christian...there was a $0$ to much in the result...sorry for that! –  dedoco Sep 26 '12 at 20:51

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.

share|improve this answer

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.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.