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I am just doing some olympiad exercises to practice my probability skills, but I have difficulties with this one:

Lets consider a village where each resident gets off work at a random time. Afterwards they go fishing. The lake only have two types fishes, $A$ and $B$. $B$ make up a proportion $p$.

This village has a law wich states that you have to stop fishing as soon as you catch the first fish of type $B$.

What proportion of all fish caught in the lake are fishes from type $B$ ?

The first thing I would do is to calculate the probability of catching fish $B$, which is equal to $p$. Now I do not how the include the other information?

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Thanks I edited it –  Alexander Mar 16 '13 at 21:32

3 Answers 3

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HINT: When a resident goes fishing, the probability that he catches $n$ fish is $p(1-p)^{n-1}$: he must first catch $n-1$ fish of type $A$ and then catch a fish of type $B$. Thus, the expected number of fish caught by that resident is

$$\sum_{n\ge 1}np(1-p)^{n-1}=p\sum_{n\ge 1}n(1-p)^{n-1}\;.$$

Consider the geometric series

$$\sum_{n\ge 0}x^n=\frac1{1-x}\tag{1}$$

for $|x|<1$. Differentiating $(1)$ gives us

$$\sum_{n\ge 1}nx^{n-1}=\frac1{(1-x)^2}\;.$$

Can you take it from here? Note that one resident’s expected proportion of fish of type $B$ is the same as the expected proportion for the entire village.

Note: If you finish the calculation correctly, you’ll find that the answer is exactly the one that you should expect intuitively.

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Ok, the expected number of fish caught by that resident is $\frac{1}{(p)}$. Therefore the proportion of caught fish of type B is p –  Alexander Mar 16 '13 at 21:51
    
@Alexander: You’ve got it. (And do look at André’s much less computational way of solving the problem.) –  Brian M. Scott Mar 16 '13 at 21:52

A statistician keeps track of all the fishers. For each person, she records the type of the first fish the person caught, the type of the second fish (if there is one) that the person caught, and so on.

Among first fish caught by residents of the village, the proportion of type $B$ fish is $p$.

Among second fish caught by residents of the village, the proportion of type $B$ fish is $p$.

Among third fish caught by residents of the village, the proportion of type $B$ fish is $p$.

And so on.

So the overall proportion of fish of type $B$ caught is $p$.

Remark: This is a fishy version of an old puzzle, about "couples" who keep having children until say the first girl, and then stop (having children). Under the usual assumptions, it turns out that the overall proportion of girls born is $\frac{1}{2}$, no change.

Another family of versions has to do with elaborate schemes for winning at, say, roulette. Same result, the more you play the more you lose.

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I think this is a rather open question, but here is my try. Let's suppose that there are $N$ people in the village. Let's call $X_i^A$ and $X_i^B$ the number of fish of types $A$ and $B$ that the person $i$ catches. Then these random variables are $$ X_i^A \sim \text{geometric}(p) ;\quad X_i^B = 1$$ Here I consider the second version of the geometric distribution, that means that $$ P(X_i^A = k) = (1-p)^k p$$ Intuitively this means that I have to not catch a $B$ fish $k$ times, and that happens with probability $(1-p)^k$ then the next one must be a $B$ fish and that happens with proba $p$. Then the proportion of fish $B$ is given by the random variable $$ Q_B = \frac{\sum_{i=1}^N X_i^B}{\sum_{i=1}^N X_i^A + \sum_{i=1}^N X_i^B} = \frac{N}{\sum_{i=1}^N X_i^A + N} = \frac{1}{\frac{1}{N}\sum_{i=1}^N X_i^A + 1}$$ Assuming independence of the $X_i^A$ we know by the law of large numbers that $$ \frac{1}{N}\sum_{i=1}^N X_i^A \to E(X_1^A) = \frac{1-p}{p} \quad \text{as } N \to \infty$$ And so, for large $N$ we can approximate the proportion as $$ Q_B \approx \frac{1}{\frac{1-p}{p} + 1} = p$$ I'm sure that there must be a cleverer argument to arrive to this.

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