I went through some books and found this problem. I am unable, and am really curious to see the proof.  There are 3 different colored balls in an urn: Red, Blue and Green.  There are 6 reds, 7 blues and 8 green.  Balls are drawn until one color is completely drawn.  Find with proof the probability that all the blue balls are drawn first.
123788
Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.   Suppose also that balls are drawn from the box one at a time at random.  What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.  The answer I was given is $\dfrac{7}{12}$ and a general equation is: $$ \dfrac{b g}{1-b}+\dfrac{b g}{1-g} $$ where $$ g=\dfrac{20}{60},b=\dfrac{30}{60} $$ but why?
32033
a) Determine the number of ways to colour a 1 by n grid using the colours red and white if an even number of squares are to be coloured red.   Note that n is greater than or equal to 1 and the number of red squares could be 0.  I know I must consider the cases of n being even and odd, and determine the sum, from zero to n of the squares being red, where the number of red squares is even. But I just want to make sure I haven’t missed any arrangements by saying a) is $2^{n-1}$ when n is even and $2^{n-1}-1$ when n is odd.  b) Determine the number of ways to colour the squares using red, white and blue if an even number of squares are to be coloured red.  For this case, I can have arrangements including red only, white only, blue only, white and blue only, red and white only, red and blue only, and finally all three colours.  Thank you for any help provided.
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.
101013
In an urn there are $n$ red balls and $m$ blue balls. I extract them without replacement. Let $X$=time of first blue. What is $E(X)$?  I found PMF of $X$ and it is, if $k &gt; n+1$  $$P(X=k) =\frac{n(n-1) \dots (n-k+1)m}{(m+n)(m+n-1) \dots (n+m-k+1)}$$  How could I evaluate $E(X)$?  Edit: I'm looking for an explicit form of $E(X)$
31215
If you have 5 marbles in a bag, 2 of them are red and 3 of them are white, what is the probability that you take out 2 red marbles out of the bag back to back? No replacement.
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.
103437
Bag A contains 3 white and 2 red balls. Bag B contains 6 white and 3 red balls. One of the two bags will be chosen at random, and then two balls will be drawn from that bag at random without replacement. What is the probability that the two balls drawn will the be the same color? Express your answer as a common fraction.
104965
A box contains $n$ identical balls numbered $1$ through $n$. Suppose $k$ balls are drawn in  succession.   (a) What is the probability that m is the largest number drawn?   (b) What is the probability that the largest number drawn is less than or equal to $m$?  I don't know how to solve this problem? Could you help me? For (a), I only know that the denominator will be $\binom{n}{k}$. How to find the numerator for both cases?
49145
Make sequence of $9$ balls from 3 red, 3 blue, 3 green, in such a way that no two balls of the same colour are next to each other. In how many different ways can you do this? (symmetric arrangements must be counted only once).  To start with, all possible arrangements are $9! = 362880$.
107198
I wanted to share a really cool but simple problem.  Consider a coloring of the points of $\mathbb R^n$ with $n$ colors. Prove that there is a color $c$ such that for any $r&gt;0$ there are two points of the color $c$ at that distance.  Regards.
5564
Given a box that contains 90% fair coins and 10% loaded coins, (a loaded coin gives heads 90% of the time), what is the probability for a randomly drawn coin to give 5 heads in a row?  P(HHHHH | random coin from box of mixed coins) = ?   I used Bayes Theorem to calculate the probability I got .219 or 21.9% probability for the random coin to give 5 heads in a row!    But I am still doubting if my answer is correct !!!  Any help appreciated :)
116341
I'd like to ask for help with a task that I've been given.  The task goes like this: In a poker game (52 cards), 5 players are given 5 cards each. What is the probability that both the first and the third player will get a full-house?  It's apparent that the probability of the first player getting a full house is  $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}}$  Then the second player gets 5 random cards, so $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5}$  Now the third player is supposed to get a full house, I'm assuming it should look like this: $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}}$  The fourth and the fifth player both receive 5 random cards, so the final equation should be:$\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}} * \binom{37}{5}* \binom{32}{5}$  This is how I would solve the task, but I'm not sure that my solution for the third player is correct. Could anyone more experienced than me have a look at my solution?  Also I'm sorry if there is a solution for this type of task on this site, I tried looking for it and I found a lot of poker probability questions, most of them were however focused on only one player and not multiple players, which is my case.
45915
How many ways are there to distribute 5 distinguishable balls between 3 kids such that each of them gets at least one ball?   My approach is $ \binom{5}{3} 3! $ + $ \binom{2}{2} \binom{3}{2}2!$ which is equal to $66$?
26788
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?
21919
I have been looking at the birthday problem (http://en.wikipedia.org/wiki/Birthday_problem) and I am trying to figure out what the probability of 3 people sharing a birthday in a room of 30 people is. (Instead of 2).  I thought I understood the problem but I guess not since I have no idea how to do it with 3.
85603
A bag contains $4$ balls .Two balls are drawn at random and are found to be white.  What is the probability that all the four balls in the bag are white?.  The options are $\large\frac{1}{5}\space ;\frac{2}{5} ;\frac{3}{5} ;\frac{4}{5}$  My work: It's given that two balls are white.So the rest of the balls could be    all white One of them white none of them white   So among the three above possibilities only one satisfies our conditions so the probability is $\frac{1}{3}$.But it not even the options given. What's wrong with my work?
62763
I was wondering if someone could critique my argument here. The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday.  My argument is that there are 23 choose 2 ways times $\displaystyle \frac{1}{365^{2}}$ for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday. This is just 365 permute 21 times $\displaystyle \frac{1}{365^{21}}$. To summarize:  $$\binom{23}{2} \frac{1}{365^2} \frac{1}{365^{21}} P\binom{365}{21}$$