In a jar of balls with different colours, can we find a general probability distribution of having n distinct colours with N number of balls grabbed ? Assuming that any colours have an equal chance to be selected. For example :  In a jar of balls with only one colour, then we only need to draw one ball (in order to see one colour).   In a jar of balls with 2 different colours, then on average we will probably need to draw 2 (or 3) balls.   In a jar of balls with 3 different colours, on average we will need to draw 5 (or 6) balls (to see 3 different colours observed).   In a jar of balls with 4 different colours, on average we need to draw 7 balls (to see 4 different colours observed).  In a jar of balls with 8 different colours, then on average we might need to draw 15 balls (to see 8 different colours observed)  In a jar of balls with 10 different colours, then on average we might need to draw 30 (or maybe 40) balls to see 10 different colours observed  In a jar of balls with 30 different colours, then we might need to draw an infinite number of balls to see the 30 different colours  and so on (need a very large number of draws if we have larger number of colours in the jar, to observe the balls with different colours)  So, by looking at this pattern, can we find the probability model for this kind of problem ?  Thanks..
21485
I'm trying to solve the well known Coupon Collector's Problem by explicitly finding the probability distribution (so far all the methods I read involve using some sort of trick). However, I'm not having much luck getting anywhere as combinatorics is not something I'm particularly good at.   The Coupon Collector's Problem is stated as:  There are $m$ different kinds of coupons to be collected from boxes. Assuming each type of coupon is equally likely to be found per box, what's the expected amount of boxes one has to buy to collect all types of coupons?  What I'm attempting:  Let $N$ be the random variable associated with the number of boxes one has to buy to find all coupons. Then $P(N=n)=\frac{|A_n|}{|\Omega _n|}$, where $A_n$ is the set of all outcomes such that all types of coupons are observed in $n$ buys, and $\Omega _n$ is the set of all the possible outcomes in $n$ buys. I think $|\Omega _n| = m^n$, but I'm not even sure about that anymore, as all my attempts so far led to garbage probabilities that either diverged or didn't sum up to 1.
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?
91150
Each of 15 red balls and 15 green balls is marked with an integer between 1 and 100 inclusive; no integer appears on more than one ball. The value of a pair of balls is the sum of the numbers on the balls. Show that there are at least two pairs consisting on one red ball and one green ball, with the same value. Show that this is not true if there are 13 balls of each colour.  since each ball has a different number and if no two pairs have the same value there is going to be $15*15$ different sums. Looking at the numbers 1 through 100 the highest sum is 199 and lowest is 3, giving 197 possible sums. Since $15*15 = 225 &gt; 197$ there must be two pairs with the same colour.   Looking at the 13 case, $13*13 = 169 &lt; 197$ and so there is not necessarily two pairs with the same sum.  My question is; is my reasoning correct, is there a more elegant proof and what category does 14 fall under?  This question is from 'An Introduction to Combinatorics and Graph Theory' and falls under the Pigeon-Hole principle and Ramsey numbers.  For the case of 14, $14*14 = 196 &lt; 197$ but which would indicate that there are not two pairs with the same sum, yet the question defaulted to 13 rather than 14 suggesting otherwise.
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If $n$ distinct objects are arranged in a circle, I need to prove that the number of ways of selecting three of these $n$ things so that no two of them are next to each other is $\frac{1}{6}n(n-4)(n-5)$.  Initially I can select $1$ object in $n$ ways. Then its neighbours cannot be selected. So I will have $n-3$ objects to choose $2$ objects from. Again, I can select the second object in $n-3$ ways. The neighbours of this object cannot be selected. However from here onwards, I am unable to extend this argument as the selection of the third object is dependent on the position of the first and the second object. Is there any simpler method to prove the result?
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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?
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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.
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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?
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I've seen many variations of this problem but I can't find a good, thorough explanation on how to solve it. I'm not just looking for a solution, but a step-by-step explanation on how to derive the solution.  So the problem at hand is:   You have m balls and n bins. Consider throwing each ball into a bin uniformly and at random.    What is the expected number of bins that are empty, in terms of m and n? What is the expected number of bins that contain exactly 1 ball, in terms of m and n?   How would I approach solving this problem?  Thanks!
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There are 50 numbered objects from 1 to 50. We extract 3 of these objects one after the other without reinsertion. Calculate the probability P(a&lt;b&lt;c)  Given 3 objects, they can be ordered in $3!$ different ways. Of all these ways only 1 satisfies a&lt;b&lt;c so the probability should be:  $P(a&lt;b&lt;c) = \frac{1}{3!} = \frac{1}{6}$  Is it solved in the right way?
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The following is a question I've made myself, but I need help in solving it:  Suppose there are 100 balls in a box. 20 balls are blue, 30 balls are green and 50 balls are yellow. Now we randomly pick out 10 balls out of the box (one ball after the other) and we don't put the balls back in the box.  What's the probability of the 10th ball being picked having color blue?  I tried thinking of all the possibilities of the the 10th ball being blue, divided by all of the possible combinations. I tried for several hours and couldn't figure out neither of them. Help is appreciated!
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If I am selecting 232 people from a pool of 363 people without replacement what is the probability of 2 of a list of 12 specific people being in that selection?  This is a random draw for an ultra race where there were 363 entrants for 232 spots. There is an argument about whether the selection was biased against a certain group of 12 people.  My initial attempt at calculating this was that there was 232 choose 363 possible selections. The number of combinations of any one person from the list of twelve is 1 choose 12 + 2 choose 12 + ... + 11 choose 12 + 12 choose 12. Thus 1 choose 12 + 2 choose 12 .... / 232 choose 363. Which ends up being a very low number which is clearly too low.  How do I calculate this?
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How many ways are there to distribute $7$ (identical) apples, $6$ oranges and $7$ pears among $3$ different people with each person getting at least $1$ pear?   Below are my workings but I am not sure if they are correct, looking for some help, thanks!  Using the inclusion/exclusion principle:  Include the number of ways to distribute them such that at most $\color\red3$ people have pears:  $$3^7\cdot\binom{3}{\color\red3}\cdot\color\red3^6=1594323$$  Exclude the number of ways to distribute them such that at most $\color\red2$ people have pears:  $$3^7\cdot\binom{3}{\color\red2}\cdot\color\red2^6=419904$$  Include the number of ways to distribute them such that at most $\color\red1$ person has pears:  $$3^7\cdot\binom{3}{\color\red1}\cdot\color\red1^6=6561$$    Hence the number of ways to distribute them such that each person has pears is:  $$1594323-419904+6561=1180980$$
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This problem is related to my lab's research in robotic coverage:  Randomly draw $n$ numbers from the set $\{1,2,\ldots,m\}$ without replacement, and sort the numbers in ascending order.  $1\le n\le m$.   From this sorted list of numbers $\{a_{(1)},a_{(2)},…,a_{(n)}\}$, generate the difference between consecutive numbers and the boundaries: $g = \{a_{(1)},a_{(2)}−a_{(1)},\ldots,a_{(n)}−a_{(n-1)},m+1-a_{(n)}\}$. This gives $n+1$ gaps.  What is the distribution of the maximum gap?   $P(\max(g) = k) = P(k;m,n) = ? $  This can be framed using : $P(g_{(n+1)} = k) = P(k;m,n) = ? $  See , but this question asks the distribution of the maximum gap.  I'd be satisfied with the average value, $\mathbb{E}[g_{(n+1)}]$.  If $n=m$ all the gaps are size 1. If $n+1 = m$ there is one gap of size $2$, and $n+1$ possible locations. The maximum gap size is $m-n+1$, and this gap can be placed before or after any of the $n$  numbers, for a total of $n+1$ possible positions. The smallest maximum gap size is $\lceil\frac{m-n}{n+1}\rceil$. Define the probability of any given combination $T= {m \choose n}^{-1}$.    I've partially solved the probability mass function as $ P(g_{(n+1)} = k) = P(k;m,n) =  \begin{cases} 0     &amp; k &lt; \lceil\frac{m-n}{n+1}\rceil\\ 1     &amp; k = \frac{m-n}{n+1} \\ 1     &amp; k = 1 \text{ (occurs when $m=n$)} \\ T(n+1)&amp; k = 2 \text{ (occurs when $m=n+1$)} \\ T(n+1)&amp; k = \frac{m-(n-1)}{n}  \\ ?     &amp;  \frac{m-(n-1)}{n} \le k \le m-n+1 \\ T(n+1)&amp; k = m-n+1\\ 0     &amp; k &gt; m-n+1 \end{cases}  \tag{1} $  Current work (1):  The equation for the first gap, $a_{(1)}$ is straightforward: $$ P(a_{(1)} = k) = P(k;m,n) = \frac{1}{{m \choose n}} \sum_{k=1}^{m-n+1} {m-k-1 \choose n-1} $$ The expected value has a simple value: $\mathbb{E}[P(a_{(1)})] =  \frac{1}{ {m \choose n}} \sum_{k=1}^{m-n+1} {m-k-1 \choose n-1} k = \frac{m-n}{1+n}$. By symmetry, I expect all $n$ gaps to have this distribution. Perhaps the solution could be found by drawing from this distribution $n$ times.  Current work (2): it is easy to run Monte Carlo simulations.  simMaxGap[m_, n_] := Max[Differences[Sort[Join[RandomSample[Range[m], n], {0, m+1}]]]]; m = 1000; n = 1; trials = 100000; SmoothHistogram[Table[simMaxGap[m, n], {trials}], Filling -&gt; Axis, Frame -&gt; {True, True, False, False}, FrameLabel -&gt; {"k (Max gap)", "Probability"}, PlotLabel -&gt; StringForm["m=``,n=``,smooth histogram of maximum map for `` trials", m, n, trials]][![enter image description here][1]][1]
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How would you solve this probability problem?     Group of men are playing game. Each rolls a dice.      If the number is 1, he must say "I like pink" regardless of his true taste and       If the number is 2, he must say " I don't like pink" regardless of his true taste.       If the number is 3,4,5 or 6, he must tell his true taste.    Assuming that 2/3 of men answered "I like pink", what is the probability that a man actually likes pink?
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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.
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I was hoping I could receive some clarification into the the four cases:   Placing labeled balls in unlabeled boxes with repetition. Placing labeled balls in unlabeled boxes without repetition. Placing unlabeled balls in unlabeled boxes with repetition. Placing unlabeled balls in unlabeled boxes without repetition.   I just want an intuitive understanding of the four cases.  How many ways are there to arrange, say, 4 balls into 9 boxes or 9 balls into 4 boxes in each case above, and why?
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Suppose 27 cubes are stacked together, suspended in the air, to form a larger cube.  The cube is then painted on all the exposed surfaces and dried.  The smaller cubes are then randomly permuted in both spatial position and spatial orientation to form another large cube.  What is the probability that new larger cube is identical to the original?  This is not homework, it was passed on by a friend who found it somewhere on the internet, and they do not remember where.