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.

I have two problems which I know how to solve now, but I am still not quite sure why my initial solutions are incorrect. I would really appreciate a thorough explanation of where I went wrong. Thank you.

Problem #1: In a pond there are 105 fish, 40 trout, 65 carp. A fisherman catches 8 fish, what is the probability of exactly two of them being trout if at least three of them are not (so they are carp)? I approached this by reducing the sample to 102 (assuming 3 carps), and counting $$\frac{\binom{40}{2} \binom{62}{3}}{\binom{102}{5}}.$$ I thought since 3 carps are already there, we are looking for P that two of the other 5 are trout, and the other 3 are carps. But this was wrong, and the correct answer was $$\frac{\frac{\binom{40}{2} \binom{65}{6}}{\binom{105}{8}}}{\frac{\sum_{x=3}^{8} \binom{40}{8-x} \binom{65}{x}}{\binom{105}{8}}}.$$ This also makes sense, but I don't understand why my original solution was wrong (it over-counted).

Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: $$\begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*}$$ I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.

You can catch two carp in $\binom{3}{2}=3$ ways; you can catch a carp and a trout in $\binom{4}{1}\binom{3}{1}=12$ ways; that's a total of $15$ ways.

You are trying to count them by saying "put one carp away; then I just need to select one more fish from the remaining 6; that's $\binom{6}{1} = 6$ ways. And the total number of catches is just one fish from the remaining ones, so that's also $6$". This is what you did in your fraction; but it doesn't give the right count: it undercounts the total number of possible catches, and the number of catches of the type you want. So you are not quite so much "overcounting", as you are undercounting the "universe" and also undercounting your desired outcomes; in the end, the two errors don't quite compensate and you end up with a smaller denominator than you should and a somewhat smaller numerator than you should, and the quotient ends up too big. (In my example, the probability would have come out to $1$, when it should be $12/21$).

In your second one, there is a problem with "hidden denominators". Your first product is okay, because the number of ways of selecting the 5 new balls is $\frac{5\times4\times 3}{3\times 2\times 1}$ (order does not matter), and the way of selecting 3 balls from 18 is $\frac{18\times 17\times 16}{3\times 2\times 1}$, so those denominators cancel; likewise with your second factor.

But in your second summand, this does not work out: when you are selecting two old and one new ball, you "really" have $\frac{8\times 7\times 10}{2}$, because the order which you select the new balls does not matter; but the denominator is still $\binom{18\times 17\times 16}{3\times 2\times 1}$; so you are short by a factor of $3$, since you have $$\frac{8\times 7\times 10}{18\times 17\times 16}$$ instead of $$\frac{\quad\frac{8\times 7\times 10}{2}\quad}{\quad\frac{18\times 17\times 16}{3\times 2\times 1}\quad} = \frac{(8\times 7\times 10)\times(3\times 2\times 1)}{2\times(18\times 17\times 16)} = \frac{8\times 7\times 10\times 3}{18\times 17\times 16}.$$ Likewise, your third summand is incorrect by a factor of $3$ for the same reason. You are taking into account order when you shouldn't, and because of the cancellation it shows up as a fraction that is too small (but not really an "undercount"; in fact, you are overcounting both the "desired outcome" and the "possible outcomes", but overcounting the latter more).

share|improve this answer
    
Thanks Arturo. But does your first comment apply even though in my problem it's exactly two trout, and the rest -- carps? I feel like your description is when we have at least two trouts. In my problem (it's worded a bit strangely) we need to catch exactly two trouts if at least 3 of 8 fish are not trouts (so they are carps). So basically it's precisely 2 trouts and 6 carps, since there is nothing in the middle. –  ZFCm Sep 25 '11 at 9:20
    
@ZFCm: No, it's still doesn't work. Look at the actual counts, and compare them with what you are doing. Remember: we are considering the carp to be distinguishable among themselves, and the same thing with the trout. You can't just set aside two spots and pretend they don't exist (which is what you are doing) because the counts don't come out right. Exactly as in the example I gave. You aren't "overcounting", your probability is off because both your counts (desired outcomes and possible outcomes) are undercounting, but you are undercounting the possible outcomes worse. –  Arturo Magidin Sep 25 '11 at 19:41
    
@ZFCm: One problem is that you are misindentifying the error: your error in the first problem is not "overcounting", and your error in the second problmem is not "undercounting": both errors are miscounts, but you are miscounting both the total outcomes and the "desired" outcomes. –  Arturo Magidin Sep 25 '11 at 20:30
    
Thanks, it makes sense now! –  ZFCm Sep 26 '11 at 19:32

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.