We have an experiment where we observe a random number of objects (samples) and we measure how many of those are defective. Imagine that this is done in different factories that they produce these objects in different speeds, so 1 factory might produce 1 object a day, and others 1000 or more. Is there any formal way (or any ideas) to produce an ordering of the factories that reflects how defective products they have (from the worse to the best)? The problem is that if we take into account only the ratio   $\frac{defectives}{observed}$ it does not produce a "fair" ordering since   $\frac{1~defective}{2~observed}=\frac{20~defective}{40~observed}$ while the second ratio should be considered worse since it has a bigger sample.
120762
Let's say I have a database of all websites and their corresponding viewers and buyers. I want to rank the websites based on the conversion rate or where we see more conversion.   So example:  website   viewer   buyer abc.com   1000     10 xyz.com   10       2 def.com   200      1   How do I rank the websites so that I know the top websites are the ones which have higher conversion rates.   Conversion is buyer/viewer
29766
Help me to solve this problem please..  Let $Y_{(1)}, Y_{(2)}, Y_{(3)}, Y_{(4)}, Y_{(5)}$ denote the order statistics of a random sample of size 5 from a distribution having p.d.f. $f(y) = e^{(-y)}, 0 &lt; y &lt; \infty$, zero elsewhere. Show that $Z_1 = Y_{(2)}$ and $Z_2 = Y_{(4)} − Y_{(2)}$  are independent. Hint: First find the joint p.d.f. of $Y_{(2)}$  and  $Y_{(4)}$.
26547
Assume the dice are numbered from $1$ to $k$. My hunch is that this will form a normal distribution with a median at $n\cdot\frac{k}{2}$. However, I have no idea as to    turn this fact into an answer (I have a minimal knowledge of stats, but I know that I am missing the standard distribution) and this is probably the wrong approach   How can I approach and solve this problem? (Aside, this is not for a class, stats or other, so any and all approaches welcome).  *Edit: * I want to find the number of ways that the sum of the numbers that are rolled has a particular value, if $n$ dice are rolled, and each has $k$ sides, numbered $1$ to $k$.
39932
Given $n$ items with weight $w_n$ each -- what is the probability that item $i$ is chosen in a $k$-out-of-$n$ "weighted random sampling without replacement" experiment? Can a closed-form solution that depends only on $w_i / w_\cdot$ be derived ($w_\cdot = \sum_j w_j$.)?   EDIT: A solution that depends only on $w_i / w_\cdot$ is impossible. Assume $n=3$, $i=1$, $w_1 = 1$, and two cases: (1) $w_2 = 1, w_3 = 1 + \varepsilon$, (2) $w_2 = 2, w_3 = \varepsilon &gt; 0$. In case (1) the probability is almost $2/3$, in case (2) it is almost $1$, but in both cases $w_1 = 1$ and $w_\cdot = 3 + \varepsilon$.  What I have tried so far to solve the problem:  Let $P^n_k(w, i)$ be the probability that item $i$ is chosen in an $k$-out-of-$n$ experiment with weight vector $w$. In the first draw, the item is chosen with probability $w_i / w_\cdot$. Otherwise, we are looking for the probability to choose this item in a $k-1$-out-of-$n-1$ experiment, with the same weight vector except for the item that has been selected. Hence:  $$ P^n_k(w, i) = w_i / w_\cdot + \sum_{j \neq i} w_j / w_\cdot \cdot P^{n-1}_{k-1}(w - j, i) $$ $$ = w_\cdot^{-1} \left( w_i + \sum_{j \neq i} w_j \cdot P^{n-1}_{k-1}(w - j, i) \right) $$  with $w-j$ being "the vector $w$ without the $j$th element".  How to solve this recurrence relation? (If it is correct at all...)
119873
Suppose that $N$ pairs of socks are put in a washing machine, with each sock having one mate. If the washing machine randomly eats socks, and at the end of the wash returns a random number $K$ of socks where $0 \leq K \leq 2N$, where each $K$ is equally probable, what is the expected number of complete pairs of returned socks?  Just from working out the first few values of $N$, I conjecture that the answer is $N/3$, but I am not sure how to prove it for all values of $N$.  Just to be clear, this is the expected value for an infinite number of trials where each $K$ is equally probable, not the expected value for an infinite number of trials where $K$ is fixed, whereupon the answer is $\displaystyle{K \choose 2}/(2N-1)$.
100540
This is a variation of a famous problem, that I have heard before but I have never seen it in this format before.      $N$ Guests arrive at a party, each wearing a hat. We collect all the hats and then randomly redistribute the hats, giving each person one of the $N$ has randomly. Letting $X_N$ be the number of hats that are matched correctly, what is the $PMF$ of $X_N$? I.e.    $P(X_N=k)=$   The common variations of this problem that I have seen usually involve the probability that no one gets the correct hat or as $N$ approaches $\infty$. I am struggling to identify a distribution that would result in any type of general formula for $PMF$ of this random variable. Has anyone seen this before?
50622
I reading my textbook and I don't understand the concept of distributions or number of solutions to an equation.  It's explained that this problem is 1/4 types of sampling/distributions problems.  An example is provided to illustrate:     In how many ways can 4 identical jobs (indistinguishable balls) be distributed among 26 members (urns) without exclusion (since one member can do multiple jobs)?      A sample outcome might be:   $\text{_____________}$   |  A  |  B  |  C  |...|  Z  |    $\text{--------------------}$      $\text{_____________}$   |  o  | oo|    |   |    ||| o  |    $\text{_____________}$   |  A  |  B  |  C  |...  |  Z  |    $\text{--------------------}$      Thus, the question is reduced to, "How many $(26-1+4)$ letter words are there consisting of four circles and $(26-1)$ vertical lines?"  Therefore, the solution is:  $\binom{26-1+4}{4}$   I really don't understand why its $(26-1+4)$.  There's only 26 different spots or "urns" to place the 4 jobs.  Can someone please explain?    I looked through another text to try and understand and I found it explained as such:       There are $\binom{n+r-1}{r-1}$ distinct nonnegative integer valued vectors $(x_1,...,x_r$ satisfying the equation $(x_1 + ... + x_r = x_n$ for $x\ge0$. $\spadesuit$   How in the world are they deriving this?  For distinct positive integers I understand:  Assume I have 8 balls (n=8) and I have 3 urns (r=3): o^o^o^o^o^o^o^o, where o represents  a ball and ^ represents a place holder where an urn could be placed.  For this scenario:     There are $\binom{n-1}{r-1}$ distinct positive integer valued vectors $(x_1,...,x_n)$ satisfying the equation:  $x_1 + ... + x_n = n, x_i&gt;0, i=1,..,r$ $\clubsuit$   It's clear that I could have this specific case ooo|ooo|oo.  Here the bar represents a divide for the urn and you see I have 3 sections.  So that case is clear.  Can anyone please explain this problem to me?  I don't understand the nonnegative integer case.   Also, people who post tend to be crazy smart and explain things very in a complicated manner.  I'd appreciate it if it could be explained in layman's terms as much as possible.   Thank you!!!
106599
If I have n samples from an $N(\mu,\sigma)$ distribution, how can I estimate the distribution from a subset of m of these n samples where the order within those n is known.  For example, if n = 100 and m is 3, I might know that one sample is 10 and 3rd highest of the 100, one is 5 and 42nd highest, and the last is 2 and 74th highest.
55887
Given there are 365 days in a year and $n$ people, the probability that at least two people have the same birthday is given by  $$\frac{365^n - 365 \cdot 364 \cdots (365-n+1)}{365^n}.$$ In this calculation, we treat the $n$ birthdays we selected as a sequence where their order matters.   If we think the $n$ dates that randomly selected as a set where the order does not matter, I believe we would have the probability $$\frac{{365+n-1 \choose n} - {365\choose n}}{{365+n-1 \choose n}}.$$   For the total number of outcomes, we model by throwing $n$ balls uniformly randomly into $365$ boxes. Essentially, there are $n$ balls and $365-1$ walls (between the boxes). For example $$ \circ |\;\;\; |\;\;\;|\circ \circ |\;\;\;|\circ |\cdots |$$ would mean one person was born on day $1$, two people were born on day $4$, one person was born on day $6$, and so on. This is how I came up with the ${365+n-1\choose n}$. For the case that everyone has a distinct birthday, it would be ${365 \choose n}$.   What is wrong with this approach?
66702
I was solving some combinatorics question and I came upon one I've been trying to solve for 2 days and I'm curios how to solve it.  It is known that in every set of 15 cars, at least 3 were manufactured from the same country (there can also be another 3 (or more) cars in the set which came from another country). prove that in every set of 100 cars, 15 were made manufactured in the same country.  Thanks to anyone who helps!
27155
Suppose we have an urn with one red ball and one blue ball. At each step, we take out a single ball from the urn and note its color; we then put that ball back into the urn, along with an additional ball of the same color.   This is Polya's urn, and one of the basic facts about it is the following: the number of red balls after $n$ draws is uniform over $\{1, \ldots, n+1\}$.   This is very surprising to me. While its not hard to show this by direct calculation, I wonder if anyone can give an intuitive/heuristic explanation why this distribution should be uniform.   There are quite a few questions on Polya's urn on math.stackexchange, but none of them seem to be asking exactly this. The closest is , where there are some nice explanations for why, assuming as above that we start with one red and one blue ball, the probability of drawing a red ball at the $k$'th step is $1/2$ for every $k$ (it follows by symmetry).
116047
I have a population with a mean (mean_1) (which was calculated through many trials and so have a very very small uncertainty in the mean). The standard deviation of the population is 'S'.  I have a second population with mean_2 (which again has a very very small uncertainty) and a standard deviation which is again 'S'.  Say I make a single further observation and the value falls somewhere in between mean_1 and mean_2, is there a test that I can perform to say with what certainty the observation came from population 1 and what certainty it came from population 2?  Thank you.
12001
Why is  even a valid operation in the first place? I mean it can give us some nice results, but why can we do the ordering for the operators like that?  Is its definition motivated by the relation between the normal ordering and the time-ordered product, which is basically the content of Wick's theorem?
57410
I was working on historical data looking for anomalous patterns that we would not expect to occur at random. I'd like to create a scheme to analyze data and markets to test for statistical significance and consistency over time.  Question : Are there some tools we could use to evaluate the nonrandomness of a database? What do we use usually to say that something is random or not (statistically)? Figure we work with change in some value over time.
13317
If I draw 5 samples from a distribution that is cont., what is the probability that the median of the population is between the xmin and xmax of the draw samples?  What is the formula to apply here?
78047
Consider the following problem:  We have a collection of m white and n black balls. We pick k balls (k ≤ m+n) without replacement. What is the expected number of white balls in the sample?  Counting method 1: The ratio of white balls in the collection is m/(m+n). If we repeatedly sample the collection, we expect that on average, our sample has the same ratio. Therefore, the expected number of white balls in a sample of size k is km/(m+n). As a side note, I think the above argument somehow depends on the unbiasedness of  the mean estimator. Am I right?  Counting method 2: We directly compute the expected number of white balls in the sample: $$ \frac{\sum_{i=0}^{k} i\binom{m}{i}\binom{n}{k-i}}{\binom{m+n}{k}} \enspace.$$  So, we reach at the identity: $$ \sum_{i=0}^{k} i\binom{m}{i}\binom{n}{k-i} = \frac{km}{m+n}\binom{m+n}{k} \enspace.$$  How to prove the above result without double counting? It seems to be provable using  as well as $i\binom{m}{i} = m\binom{m-1}{i-1}$. But is there a simpler approach (which does not depend on extra identities such as Vandermonde's)?
58442
I am looking at this formula  My understanding is that the number of partitions is equal to that formula, but only when the order of the partitions matters. eg, That's right, if you can only have $n_1$ in the first group and $n_2$ in the second group and so on.   My understanding is that if the order of the groups does not matter, then it is the same formula except you add $r!$ to the denominator because you need to account for the different orderings that can achieve that result.  Is that right?