Suppose I buy a big box of assorted sweets. There are 10 distinct types of sweets, all evenly distributed in their assortment. I am separating them out into party bags for a party. I make up 100 party bags, each containing four distinct sweets. As such there are 400 sweets in total, with 40 of each variety distributed amongst the bags.  The question is, if I look inside $n$ bags (without replacement), what is the probability that I will have seen every variety of sweet?   I appreciate that if I knew the exact tally of sweet varieties I wanted to find I could use a multivariate hypergeometric distribution. This is a version of the coupon collecting problem but without replacement. It may be that this problem isn't easily analytically solvable? Am I doomed to be unable to do this for such a large dimension of variables? Any advice or input would be greatly appreciated. Perhaps calculating a Chao index is required?
38673
I was looking at various extensions of the classical Coupon Collector problem, but couldn't find any answers or hints for the following modification:  Assume there are $n$ distinct coupons and you get them in batches of a different (random) size. That is, assume that every time you get a black box with coupons in it, whose number can be from $0$ to $n$ and coupons within a black box are distinct. The probability of having a specific coupon in a box is $p$ (that is, it's the same for all). What is the expected number of black-boxes one needs to open to collect all coupons?
30682
This is my reconstruction of the EdExcel GCSE question that has caused such a  in the UK in the last 24 hours, along with its solution.    Hannah has a bag containing $n$ sweets, 6 of which are orange. She eats two sweets at random from the bag.  The probability that the two sweets Hannah eats are both orange is $\frac{1}{3}$. Show that $n^2 - n = 90$.    Possible follow-up (I don't know if this was part of the original): how many sweets were there in total in the bag?
17758
I understand how to calculate multivariate hypergeometric distributions.  For example, if you have an urn with 2 red marbles, 4 white marbles, 8 blue marbles, and 12 orange marbles, the probability of drawing 5 marbles and getting 1 red marble and 2 white marbles is as follows:  $P = \frac{{2 \choose 1}{4 \choose 2}{20 \choose 2}}{26 \choose 5}$  What I'm wondering is if I wanted to calculate 1 or more red marbles and 2 or more white marbles, if that could be calculated this way:  $P = \frac{{2 \choose 1}{4 \choose 2}{24 \choose 2}}{26 \choose 5}$  where I'm adding in the extra red and white marbles to the last combination. If this doesn't work (which I don't think it does because the numerator and denominator no longer match) why not, and is there a nice solution to this problem that doesn't involve summing all of the separate probabilities?
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...)
93686
I am taking a probability course and we have been of course learning the classic problems of men with hats or variations of that in some other form. The problems being of the type if $n$ men have $n$ hats and mix them all together, whats the probability $0, 1, 2,$ or $k$ men choose their hat.   But from what I can gather is that they are all problems of the same type, for any given random permutation of $k$ objects how many fixed points are there. My professor taught us to solve this using conditional probabilities (conditioning on whether the first person matched and then generating a system of equations) and using the inclusion-exclusion principle. But is there any other way to reason this?
10845
Suppose there are $m$ red balls and $n$ blue balls in an urn. We randomly choose $p:m&lt;p&lt;n$ balls uniformly from the urn. What is the probability that exactly $q$ red balls are chosen?  Note:- Normally the answer would be $\frac{{m}\choose{q}}{{m+n}\choose{q}}$. However, since the number of balls that are chosen are provided, it is confusing me out. Any hints for the answer will also be appreciated.
82736
This question is generalization of different cases of  combinatorics problems that are generally asked.  We will find general way of arranging $n$ balls in $r$ boxes. Cases :    Identical Balls. Distinct Balls. Order considered. Distinct Balls. Order not considered.   Two cases arise in each case :    Empty boxes allowed. Empty boxes not allowed.   Please feel free to add any other cases or minor variations you might think can be generalized.
84038
Given $20$ people, what is the probability that, among the $12$ months   in the year, there are $4$ months containing exactly $2$ birthdays and   $4$ months containing exactly $3$ birthdays?   I am less concerned about knowing the answer than making sure I have thought about the problem correctly.   The general problem seems to be finding the distinct integer-valued vectors that satisfy $$x_1 + x_2 + ... x_{12} = 20\ \ \ \ \ \ \ \ \ x_i \ge 0, i = 1, ..., 12$$  It can be proven that there are $\binom{20+12-1}{12-1}$ such vectors. I am presuming this to be the sample space.  (Please correct me if I'm wrong)  $4\cdot 2 + 4\cdot 3 + 4 \cdot 0= 20$, so there are $3$ distinct groups of $4$ months. Therefore we have $\binom{12}{4\ 4\ 4}$ possible ways to get the desired distribution of months.   The probability I get is $$\frac{\binom{12}{4\ 4\ 4}}{\binom{20+12-1}{12-1}} $$. The answer in the back of the textbook is $1.0604 \times 10^{-3}$ so my answer is off. I just want to know where I went wrong.
63006
The question asks us to classify all distributions in $\mathbb{R}^2$, say $u$, such that $$(xy)u(x,y) = (x^2-y^2)u(x,y)= 0.$$ I am genuinely stuck, any help is extremely appreciated.
75151
The problem is quite simple to formulate. If you have a large group of people (n > 365), and their birthdays are uniformly distributed over the year (365 days), what's the probability that every day of the year is someone's birthday?  I am thinking that the problem should be equivalent to finding the number of ways to place n unlabeled balls into k labeled boxes, such that all boxes are non-empty, but C((n-k)+k-1, (n-k))/C(n+k-1, n) (C(n,k) being the binomial coefficient) does not yield the correct answer.
82240
We have $n$ distinct balls ($n&gt;7$) and want to randomly (and independently) distribute them into $N$ distinct boxes ($N&gt;n$) which are placed one next to the other. a) What is the probability that all balls are placed in consecutive boxes? b) What is the probability that all balls are placed in consecutive boxes and balls with number $1$, $4$ and $7$ are also placed in consecutive boxes?  I am not sure but I will try: a) There is a block of n balls which must be placed in one of the $N-n+1$ gaps formed if we arrange the boxes and fill $n$ consecutive of them. So there are $N-n+1$ places place this block. The $n$ balls in the block can be arranged in $n!$ ways. Also the N boxes can be arranged in $N!$ ways. So all in all we have $n!N!(N-n+1)$. b) Again we have $N!$ ways to arrange the boxes but the balls can be arranged in $(n-3)!$ ways, right? I don't know what is the total number of ways to place all balls in the boxes - plus that I see no restriction regarding the number of balls per box. Any help?
65424
I know there's no uniform distribution for a countably infinite set, but I'm wondering if there's still a way to determine the probability of picking from a subset of a countably infinite set.  For example, what's the probability that I pick an odd number from the set of naturals, assuming I'm picking randomly? Is this even a coherent question? If so, is there a textbook approach to this problem in measure theory/probability theory/probability measure?  Thanks!
44208
Suppose you have a series of n trials, where the probability of success  in each trial is p.  The distribution of the number of successful trials follows a Binomial distribution with parameters (n, p).  The mean is given by np whereas the variance is np(1-p).  So far so good: this is pretty mundane Stats 101 stuff.  But suppose now that I only knew about the successful trials, and had no knowledge of the total number n of trials, which is the variable I am interested in estimating.  For example, I knew I had 100 successful trials, where each trial had a 0.1 chance of success.  Is there a known probability distribution that describes the probable outcomes for n, the total number of trials?  Estimating the mean is easy: if m is the number of successes, then it's just m/p.  But what about variance and other measures?  What if each success had a different (but known) chance of success?  Suppose I had the following records:   success1 (with p=0.1) success2 (with p=0.1) success3 (with p=0.2)   Again, a good estimation of the total number of trials can be obtained by simply summing 1/p for each successful trial.  In this case that number  is 10+10+5=25.  But what about variance and other measures?
43369
During a period of $n$ days, two persons drink beers, one each. There are $n$ different beers $B_1, B_2, B_3,\ldots, B_n$, where $n\geqslant 1$ is an integer. t Person1 drinks the beers in order, i.e., on the $i$-th day, she drinks beer $B_i$. Person2 takes a uniformly random permutation $a_1, a_2, \ldots, a_n$ of the set $\{1, 2,\ldots, n\}$ and drinks beer $B_{a_i}$ on the $i$-th day. Let $X$ be the random variable whose value is the number of days during which person 1 and person 2 drink the same beer. Determine the expected value $\mathbb E(X)$ of $X$.  I know that I should let $X$ be the number of days during which both persons drink the same beer. And  $$ X_i = \left\{\begin{array}{rc} 1,&amp;\text{if they drink the same beer on the }i\text{-th day} \\ 0,&amp;\text{if they drink different beers on the }i\text{-th day}\end{array}\right. $$  Then  $$\begin{align} \mathbb E[X_i] &amp;= 1\cdot P(X=1) = P(\text{Person 1 and Person 2 drink beer }i) \\&amp;= P(\text{Person1 drinks beer }i)P(\text{Person 2 drinks beer }i) \\ &amp;= 1\cdot P(\text{Person 2 drinks beer }i) = \frac{1}{(n-i)!} \end{align}$$   So, is my approach right? If it's will the next step be: $\mathbb E(X) = n\cdot\frac{1}{(n-i)!}$?
116389
I've encountered a problem, wchich sounds like this:  Mike has 25 roses and 5 friends. He wants to give the roses to his friends, in a way, that no friend is left without a rose, and no friend gets more than 7 roses. How many ways are there to distribute the roses among his friends? All roses are indistinguishable  I've managed to write this problem down as $$x_1 + x_2 + x_3 + x_4 + x_5 = 25 $$ where $$ 1 \leq x_1, x_2,x_3,x_4,x_5 \leq 7$$  I struggle to approach the problem
88502
Question  Suppose we have a multinomial distribution with $N$ possible outcomes, with probabilities $p_1,\ldots,p_N$. We sample this $n$ times, and denote the observed frequency of the $i$th outcome as $\xi_i$. In [1] the author claims that the distribution of the $\xi_i$ in the limit of large $n$ is:  $$p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right).\;\;\;\;\;(1)$$  We can see immediately that this must be an approximation, as it assigns nonzero probabilities for $\xi_1+\cdots+\xi_N&gt;1$. However we can see that these have vanishing probability in the limit $n\rightarrow\infty$. My question is how do we derive (1) from the multinomial distribution, and show that they match in the $n\rightarrow\infty$ limit?  My thoughts  My first thought would be to appeal to the central limit theorem. The multinomial distribution has mean $\mu_i=p_i$ and covariance matrix $\Sigma_{ij}=\delta_{ij}p_i-p_ip_j$, so we would expect this in the large $n$ limit to be described by a multivariate Gaussian with mean $\mu$ and covariance $\frac{1}{n}\Sigma$. However, things are complicated by the fact that the multinomial covariance is singular (since $\xi_N$ is determined by the other $\xi_i$s), and so the multivariate Gaussian is not defined.  To address this, we may try and consider only the first $\xi_1,\ldots,\xi_{N-1}$, which have a non-singular covariance matrix and hence well-defined multivariate Gaussian distribution. Let's take the Binomial distribution $N=2$. The frequency $\xi_1$, this has mean $p_1$ and variance $p_1(1-p_1)$, so this would be described the the Gaussian: $$\propto\exp\left(-\frac{n}{2}\frac{(\xi_1-p_1)^2}{p_1(1-p_1)}\right).\;\;\;\;\;(2)$$ The expression (1) gives: $$\propto\exp\left(-\frac{n}{2}\left(\frac{(\xi_1-p_1)^2}{p_1}+\frac{(\xi_2-p_2)^2}{p_2}\right)\right).\;\;\;\;\;(3)$$ If we substitute $\xi_2\rightarrow 1-\xi_1$, $p_2\rightarrow 1-p_1$ into (3), we can verify that this gives the same answer as (2). I have verified that this also works for $N=4$.   I'm sure that if I just bashed out the algebra for general $N$ we would get agreement between the central limit theorem and (1) when we restrict the latter to $\xi_1+\cdots+\xi_N=1,p_1+\cdots+p_N=1$. However, how can we start with the multinomial distribution and derive (1) as a limit which is valid everywhere? One idea would be to say that (1) goes to zero as $n\rightarrow\infty$ when you are not on that plane, however I am a bit uncomfortable with this as it goes to zero everywhere except the mean as $n\rightarrow\infty$, so I don't know if that argument is good enough.  [1] Wootters, William K. "Statistical distance and Hilbert space." Physical Review D 23.2 (1981): 357.
7424
In how many ways can we choose k items from n distinct items put in a circle.   Then further extension of the question, if I want to find out the number of ways to choose k items from n distinct items in a circle such that no 2 are together.  Thanks