Suppose that I have $n$ balls which I'm going to throw randomly in $k$ bins. I don't want that my bins have too many balls - say at most they can hold $C$ balls. What number of bins $k$ will allow us to say that each bin will have less than $C$ balls with probability $1 - \delta$.
120867
Let us fix a number of urns $n$ and a fixed capacity $c$. I would like to know which is the probability that $m$ balls, thrown at random in $n$ urns, "overflow", in the sense that at least one urn has assigned $\geq c$ balls.  There are several results in the literature about the maximum number of balls in an urn (in particular, Raab and Steeger's 1999 paper), but the results are all asymptotic with high probability. I need a precise analytical result.
47902
How do I calculate the number of options for $k$ different balls in $n$ cells, with at least $1$ ball in each cell?   Lets say $6$ different balls in $3$ different cells, with at least $1$ ball in each cell.   like this:  $$x_1+x_2+x_3 = 6$$  I would like to see an explanation please and not just the answer.  Thanks!
90901
Suppose we want to prove $$ k \binom{n}{k} = n \binom{n-1}{k-1}$$   In the LHS we are choosing a team of $k$ players from $n$ players. Then we are choosing a captain. In the RHS we are choosing a captain from the $n$ players. Then we are choosing the remaining $k-1$ players from the $n-1$ players.   Is this a correct interpretation?
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.
89342
Say we have Tom and John, each tosses a fair coin $n$ times. What is the probability that they get same number of heads?    I tried to do it this way: individually, the probability of getting $k$ heads for each is equal to $$\binom{n}{k} \Big(\frac12\Big)^n.$$ So, we can do $$\sum^{n}_{k=0} \left( \binom{n}{k} \Big(\frac12\Big)^n \cdot \binom{n}{k}\Big(\frac12\Big)^n \right)$$ which results into something very ugly. This ugly thing is equal to the 'simple' answer in the back of the book: $\binom{2n}{n}\left(\frac12\right)^{2n},$ but the equality was verified by WolframAlpha -- it's not obvious when you look at it. So I think there's a much easier way to solve this, can someone point it out? Thanks.
96501
Ten fish are caught in a lake, marked, and then returned to the lake. Two days later 20 fish are again caught, 2 of which have been marked.  (a) Find the probability of 2 of the 20 fish being marked if the lake has $k$ fish (assuming the fish are caught at random). (b) What value of k maximizes the probability?    Progress:     I got the (a), $$\binom{10}{2}\binom{k-10}{18}\bigg/ \binom{k}{20}$$ But how do I get part (b) where $k$ is to maximize the probability?
22567
Assume a coin has a probability p to get a head H. Suppose a coin is tossed until the partern T,T appear in the last 2 tosses. Once he got T,T then the game is finished. What is the expected number of flips is expected for the game?  I tried to find out the distribution of $N$ where $N$ stands for the expected value of the no. of tosses before getting T,T, but i don't know what kind of distribution is that, seems need another apporaches
48830
I have $k_1$ red balls and $k_2$ blue balls in a box. Let $p_r$ be probability that randomly select $r$ balls in the box without replacement, $\sum_{r=1}^{m} p_r=1$. A trial is succeeds if the trial can draw at least one red ball in the trial. Find expected number of successful trail in $N$ times.   This is my solution  Let $X$ be event that draw $r$ ball in the box which has at least one red ball.  $$P(X)=p(r) \times\sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$  Finally, we got the expected value as $$E(X)= N \times \sum_{r=1}^{m} p(r) \sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$  Is it right?
93423
Consider an urn with $k$ balls, labeled $1$ through $k$. One step shall be defined as the following procedure: randomly drawing one ball at equal probability for all balls in the urn, removing all balls with labels showing numbers greater than the number of the ball drawn, putting the ball drawn back. (Side remark: it is possible that the number of balls doesn't change by doing a step).  Question: What is the expected value for the number of steps $n$ needed to draw the ball with label $1$?  Answer (so far): If we look at the problem via tree diagrams we can conclude, that the cumulant probability to draw the ball labeled $1$ within $m+1$ steps is, starting with $k$ balls is given by $S(m,k)/k$ with:    $$S(m,k_0)=\sum_{k_1=1}^{k_0}\cdots\sum_{k_m=1}^{k_{m-1}}(\prod_{j=1}^{m}k_j)^{-1}.$$     Replacing $1/i$ with $\int_0^1dx\;x^{i-1}$ the arising terminating geometric series $\sum_{i=0}^{k-1}x^i$ with $\frac{1-x^k}{1-x}$ and integrating out $\int_0^xdy\;(1-y)^{-1}$ this can be simplified to   $$S(n+1,k)\cdot n!=\int_0^1\frac{1-x^k}{1-x}\ln^n(\frac1{1-x}),$$  where of course the word 'simplified' is up to personal preference. This form only remains valid for nonnegative $n$ and as the number of steps is given by $n+2$ we need to know in addition to the above, that $S(0,k)=1$ (ie. that the probability of finding $1$ in the first step is $1/k$).  From all that we can in principle find the Expactation value by noting, that for discrete random variables with cumulative $F(n)$ the mean can be defined as $E=\sum_{n=0}^\infty \bigl(1-F(n)\bigr)$.  Unfortunately I'm not able enough to calculate this sum. $F$ relates to $S$ by $F(n)=S(n-1,k)/k$.    EDIT: The answer provided by WhatsUp is correct. I was unsatisfied that the formula $E_m = 1+\frac1m\sum_{i=1}^mE_i$ just came up from nowhere and derived it myself.  From tree diagrams one sees that the probability $P_k(n)$ to find $1$ in the $n$-th step with $k$ balls before the first step is given by $1/k$ if $n=1$ and $P_k(n+1)=\frac1k\sum_{k_1=2}^k\cdots\sum_{k_n=2}^{k_{n-1}}\prod_{j=1}^n\frac1{k_j}$ otherwise. This directly gives $P_k(n)=\frac1k\sum_{i=2}^kP_i(n-1)$.  Inserting that into the mean formula $E_k=\sum_{n=1}^\infty n\cdot P_k(n)$ gives $$\frac1k+\frac1k\sum_{n=1}^\infty(n+1)\sum_{i=2}^kP_i(n)\\ =\frac1k + \frac1k\sum_{i=2}^k\sum_{n=1}^\infty n\cdot P_i(n) + \sum_{n=2}^\infty\frac1k\sum_{i=2}^kP_i(n-1).$$ This can be found to be equal to $1+\sum_{i=2}^kE_i/k$ by noting that the last summand has to amount to $1-P_k(1)$.    Of course then $E_k=\frac1{k-1}(k+\sum_{i=2}^kE_i)$ follows like outlined by WhatsUp and $E_k=1+\sum_{i=1}^{k-1}\frac1i$ can be shown by induction.
85581
Find the number of ways in distributing $8$ distinct balls into $6$ distinct boxes such that there is at least $1$ ball in each box.  We are well acquainted with the traditional Inclusion-Exclusion principle in solving this and it will come up as $191520$ ways.  I tried in this way:  When we distribute $8$ balls into $6$ boxes we have two possible cases  Case $1.$ There are exactly $2$ boxes with $2$ balls each  Case $2.$ There is exactly $1$ Box with $3$ balls in it.  For Case $1$, the distribution is done by choosing $2$ boxes from $6$ boxes and $2$ balls from $8$ balls and those two balls can be arranged in $2!$ ways. But I require another two balls from remaining $6$ balls to be placed in each of these two boxes which can be done in $\binom{6}{2} \times 2!$ ways. Finally remaining $4$ balls can be distributed in remaining $4$ boxes in $4!$ ways. So total number of ways for Case $1.$ is  $$\binom{6}{2} \times \binom{8}{2} \times 2! \times \binom{6}{2} \times 2! \times 4!=604800$$  For Case $2$, the number of ways is  $$\binom{6}{1} \times \binom{8}{3} \times 5!=40320$$  I am pretty sure the value obtained for Case $2.$ is correct. but I could not find where I went wrong in Case $1.$
108025
If nine coins are tossed, what is the probability that the number of heads is even?  So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.  We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$  $n = 9, k = 0$  $$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$   $n = 9, k = 2$  $$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$   $n = 9, k = 4$ $$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$  $n = 9, k = 6$  $$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$  $n = 9, k = 8$  $$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$  Add all of these up:   $$=.64$$ so there's a 64% chance of probability?
95642
Given a set $\mathcal{S}$ of size $n$, let a $k$-partition of $\mathcal{S}$ be a grouping into $k$ disjoint classes $$(S_1 ,S_2,...,S_k)$$   Where the $S_i$ do not necessarily contain the same number of elements.  Let $C(S_1, S_2, ... S_k)$ be the set of combinations which can be made by taking exactly one element of each class, (i.e. $C(S_1, S_2, ... S_k)$ is the Cartesian Product $S_1 \times S_2 \times ...\times S_k$). For a given $(n,k)$, what is the minimum number of k-partitions $N$ such that  $\bigcup\limits_{i=1}^{N} C(S_1^{(i)}, S_2^{(i)}, ...,S_3^{(i)}) = $ All Possible Combinations  Example:  for $n=4$ and $k=2$, $\mathcal{S} = \{1,2,3,4\}$ we can choose:  $$ S^{(1)} = \left\{\left\{ 1, 2\right\}, \left\{ 3, 4\right\}\right\}$$ $$ S^{(2)} = \left\{\left\{ 1, 3\right\}, \left\{ 2, 4\right\}\right\}$$  Then $$C_{1} = \left\{\left\{ 1, 3\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 2, 4\right\}\right\}$$ $$C_{2} = \left\{\left\{ 1, 2\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 3, 4\right\}\right\}$$  and $C_1 \cup C_2 =$ All possible combinations and $N=2$.
42849
We roll a die until each side appears at least once, then we stop.     What is the probability of rolling exactly $n$ dice?   I guess the answer is  $$6-6\left(\dfrac{5}{6}\right)^n\;,$$  but this may be wrong.  In trying to solve this problem, I read , but I couldn't solve the problem.
8995
There are $n$ types of balls in an urn; $a$ balls of type $1$ , $b$ balls of type $2$ , $c$ balls of type $3$ and so on. Now balls are drawn until a ball of type $1$ is obtained with condition that if a ball of type $i$ is drawn at any moment ( $i \neq 1$ ) then all balls of type $i$ will be removed from the urn. What is the expected number of draws done? (In one draw, all balls of that particular type are removed)  My approach for $n = 3$ is like:  expectation value = $a/(a+b+c)+2(ba/(a+b+c)(a+c)+ca/(a+b+c)(a+b))+3(bca/(a+b+c)(a+c)a+cba/(a+b+c)(a+b)a)$  But it's messy , is there any general form for n type of balls ?
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
122439
I have 4 unlabeled (identical) blue balls and 2 identical redz balls and 5 different (distinct) boxes. In how many ways all balls can be placed in these 5 boxes?
113008
Four dice are thrown, what's the probability that:  a) None of them fall higher than three?  b) None of them fall higher than four?  c) That four is the highest number thrown?  So for a, I wanna think about the denominator first. There are 6 possible outcomes for each dice, and we have four dice. So we're technically picking r out of n objects, aka picking 4 possible values out of 6. So the denominator/sample space should be $6^4$ right?  Now this is where I'm getting tripped up. Order technically shouldn't count, because we only care about the quadruplets that don't have a value higher than 3. I might be wrong, maybe I am, but could someone explain a bit more? My professor stated that typically in the sample space/denominator, we want order to count.   So I'm trying to think of the numerator now, so we want to find the probability that no values appear higher than 3, so some events that can occur are: $(1, 1, 2, 3), (1, 2, 2, 3), (1, 1, 1, 1)$ etc. Again, I can't see why order should count here, because if we only are concerned about what appears rather than how they appear, then we can say $(1, 1, 2, 3)$ is equal to $(1, 2, 1, 3)$? So if order does not count, and we have replacement, then does this lead the case where we use $$\binom{n-1+r}{r}$$to find out the probability?   I'm pretty sure if I can do a), I could prob do b), but if someone could maybe lead me in the right direction for c) I would appreciate that too!