Drawing from an Urn

Let there be $N$ colors with $M$ balls each one. By counting, the probability of drawing two balls, without replacement, prior to the adding of the other $14$ is $\frac{(NM)(M-1)}{(NM)(NM-1)}$ , because there are $(NM)(NM-1)$ options for drawing two balls without replacement and for each of the $NM$ balls, other $M-1$ are of the same color. Similarly, after adding the $14$ balls, there are $(NM+14)(NM+13)=(NM)(NM-1)+14(2NM+13)$ , ways of picking two balls and $NM(M-1)+14*13$ ways of picking two balls of the same color. So the probability is $\frac{NM(M-1)+14*13}{(NM)(NM-1)+14(2NM+13)}$ equal to $\frac{(NM)(M-1)}{(NM)(NM-1)}$ .

Why expressing it like that? Because if $\frac{a}{b}=\frac{c}{d}$ then $\frac{a}{b}=\frac{a+c}{b+d}$ , and similar expressions. So, $\frac{(NM)(M-1)}{(NM)(NM-1)}=\frac{14*13}{14(2NM+13)}$ . Therefore $\frac{M-1}{NM-1} =\frac{13}{2NM+13}$ . Thus, $(M-1)(2M+13)=13(NM-1)\Rightarrow M=\frac{15N-13}{2N}$ . As $M$ is an integer, then $N|15N-13$ , so $N|-13$ , and as $N$ is a positive integer, then $N=1$ or $N=13$ . If $N=1\Rightarrow M=1\Rightarrow NM-1=0$ , and we divide by $0$ so it can't be. (Also, the initial probability would be 0 and the last one would be 13/15). If $N=13\Rightarrow M=7$ . This makes sense in the probabilities, and therefore the number of balls in the urn is $NM=7*13=91$ initially.

Let the probability of grabbing a set of matching balls, before adding any new balls, be $p$ . After adding the 14 new balls, drawing two balls can result in either 2 old balls being drawn or at least 1 new ball being drawn. The probability of scoring a pair in the first case [by definition] is $p$ , and since the weighted average of the probabilities [given] is also $p$ , the probability of getting a matched pair of balls in the second case is also $p$ . Equating this probability with the definition of $p$ gives the following equation: $p = \frac{\dbinom{14}{2}}{14ab + \dbinom{14}{2}} = \frac{a\dbinom{b}{2}}{\dbinom{ab}{2}}$ where $a, b$ are the number of colors and the number of balls of each color, respectively.

Simplifying, $\frac{14\cdot13}{14\cdot2ab+14\cdot13}= \frac{a(b)(b-1)}{(ab)(ab-1)} \\ \frac{13}{2ab+13} = \frac{b-1}{ab-1}$ Rearranging and Grouping, $13ab-13 = 2ab(b-1)+13(b-1)\\ ab(13-2(b-1)) = 13(b-1+1)\\ ab(15-2b) = 13b\\ a(15-2b) = 13$

Factoring, $13 = 1 \cdot 13\\ \implies (a,b) = (1, 1) \text{ or } (13, 7)$ Since $ab>1$ , only the second solution is valid; $(a, b) = (13, 7)$ for a total of $a \cdot b = \boxed{91}$ balls.

Let n = the number of colors initially and let x = the number of balls for each color initially.

Then the probability of picking two balls of the same color initially is n * (x/(nx)) * (x-1)/(nx-1) = (x-1)/(nx-1)

The probability of picking two balls of the same color afterwards is n*(x/(nx+14))((x-1)/(nx+13)) + (14/(nx+14))(13/(nx+13))

Let nx = a. We then have the equation.

a/(a+14) * (x-1)/(a+13) + 182/((a+14)(a+13)) = (x-1)/(a-1)

Multiply by (a-1)(a+14)(a+13) on both sides to get

(x-1)(a^2 - a) + 182(a-1) = (a+13)(a+14)(x-1) 182(a-1) = (28a + 182)(x-1) 13(a-1) = (2a + 13)(x-1) 13a - 13 = 2ax - 2a + 13x - 13 (15 - 2x)a = 13x (15 - 2x)n = 13

Since 15 - 2x and n are positive integers, we test n = 13 or n = 1. n = 1, x = 1 does not work (for lots of reasons). n = 13 yields x = 7.

nx = 91. So the total number of balls is 91.

In order to solve this problem, we're going to write expressions for the probability of the urn before and after the 14 new balls are added. Then, by setting them to be equal to each other we can find a solution to the problem.

First, we'll write an equation for the original conditions, using $c$ for the number of colors and $n$ for the number of balls per color. Assuming we pick a random ball, the chance of the next ball being the same color will be the number of balls of that color minus the one just pulled, $n-1$ , over the total number of balls minus the one just pulled, $cn-1$ .

$\frac {n-1}{cn-1}$ =

Now we'll calculate the odds for the urn with the extra balls in 2 parts, pulling an original ball and pulling a new ball. The chance of pulling an original ball is the number of original balls, $cn$ , over the new total, $cn+14$ . The number of balls that can match it will be $n-1$ as before, and we'll but that over the new total $(cn-1)+14$ , which simplifies to $cn+13$ . Multiplying the two fractions will result in the chance of picking an original ball and another that matches.

$\frac {n-1}{cn-1}$ = $\frac {cn}{cn+14}$ $\frac {n-1}{cn+13}$ +

To calculate the chance of pulling a new ball, we'll use the number of new balls, 14, over the total, $cn+14$ . We'll multiply this by the chance of pulling one of the 13 remaining balls of that color. This will be the remaining balls, 13, over the total, $cn+13$ . After we multiply these two fractions, we add them to the previous one. By adding the chances of matching an original ball and the chances of matching a new ball, we have an expression for the probability of pulling any two matching balls.

$\frac {n-1}{cn-1}$ = $\frac {cn}{cn+14}$ $\frac {n-1}{cn+13}$ + $\frac {14}{cn+14}$ $\frac {13}{cn+13}$

This can be simplified by adding the fractions.

$\frac {n-1}{cn-1}$ = $\frac {(cn)(n-1)+182}{(cn+14)(cn+13)}$

We'll continue to simplify by multiplying both sides by the denominators of both sides, $(cn-1)(cn+14)(cn+13)$ , and then simplifying.

$(n-1)(cn+14)(cn+13)$ = $(cn-1)(cn^2-cn+182)$

Then, distribute fully on both sides and simplify. The final result is this:

$105c = 14cn+91$

At this point, we'll start testing integers for c until we find a solution for which n is also a positive integer. Once we try 13, we'll find $n=7$ .

$105(13) = 14(13)n+91$

$1365 = 182n+91$

$1274 = 182n$

$7 = n$

Now we know that there were originally 13 colors, with 7 balls each. Multiplying 13 and 7 results in 91, the number of balls originally in the urn and the correct answer. If we want to double check this answer, we can substitute 13 and 7 back into our full equation. Doing this will confirm both have equal probabilities of $\frac {1}{15}$ .

I made a stupid mistake formatting the solution, the final expression is c(2n-15)=13 instead of c(2n-15)=26. I apologize.

Let us assume there are n balls in the urn ,q different colors and m total sets of different colors. i.e. n = qm

Initially, as we have n choices for 1st ball , but after 1st ball is drawn of from some set of same color, that set will have m-1 balls left and total balls then remaining will be n-1. Thus, probability of drawing 2 balls of same color = \frac {m-1}{n-1}

In the case where we add 14 balls of same color (different from the ones in the urn) ,the total number of balls become n+14. Now, in the urn, we have q sets with m balls each having same color + these 14 new balls. Now we have to draw 2 balls of same color from here. Here there are 2 cases, either both come from the original balls, or both come from the new balls. If both have to come from new 14 balls, then probability will be calculated as \frac {14}{n+14} \times \frac {13}{n+13} If both balls haveto come from the original n balls, then probability will be calculated as \frac {n}{n+14} \times \frac {m-1}{n+13} ( since after taking 1 ball from those n, we actually are taking 1 ball from one of those q sets , hence for the second take, we will have m-1 options)

Thus our required probability combining both cases is ( \frac {14}{n+14} \times \frac {13}{n+13} ) + ( \frac {n}{n+14} \times \frac {m-1}{n+13} )

We are given :

\frac {m-1}{n-1} = ( \frac {14}{n+14} \times \frac {13}{n+13} ) + ( \frac {n}{n+14} \times \frac {m-1}{n+13} )

Easy simplification gives us : n = \frac {13m}(15- 2m}

We have to keep 2 things in mind now. 1) Both m and n are natural numbers >1. 2) n/m i.e. q, should also be a natural number.

The above cases are easily satisfied when denominator = 1, i.e. 15- 2m = 1 or m =7

thus n = 91 , and hence our answer.

Let there be $a$ colors and $b$ balls of each color. Remark that the probability initially is : $\frac{a \cdot b(b-1)}{ab(ab-1)}$ After adding in $14$ , it becomes: $\frac{a \cdot b(b-1) + 14(13)}{(ab+14)(ab+13)}$ Thus: $\frac{a \cdot b(b-1)}{ab(ab-1)} = \frac{a \cdot b(b-1) + 14(13)}{(ab+14)(ab+13)}$ Some simple (but tedious) algebra shows that this reduces to: $\frac{b(2ab - 15a + 13)}{(ab-1)(ab+13)(ab+14)} = 0 \Longleftrightarrow 2ab - 15a + 13 = 0$ Clearly $b > 1$ . Now, note that $b < 8$ as otherwise $2ab + 13 > 15a$ . Checking $2 \le b \le 7$ we get $b = 7, a = 13$ is the only integer solution and thus the answer is $13 \cdot 7 = \boxed{91}$ .

Let there be x colors, and n balls of each color initially. So, using the data, P1= \frac {x.{n \choose 2}}{n.x \choose 2}, which is equal to the P in the final situation, where, P2= \frac {x.{n \choose 2} + 14 \choose 2}{n.x + 14 \choose 2} Simplifying it, we get, x= \frac {13}{15 - 2.n} Using some logic and the fact that x, n are integers. We get n=7 and x=13. Since, x is not equal to 1 because then P1 cannot be equal to P2. Thus number of balls is n.x=91.

let $n=a\times b$ , where $n$ is the number of balls , $a$ is the kind of color and $b$ is the number of balls in each color.

in the firsrt case,there are ${n \choose 2}$ to choose 2 ball without replacement, and $a\times {b \choose 2}+{14\choose 2}$ to choose 2 ball with the same color.

ih the second case, there are ${n+14 \choose 2}$ to choose 2 ball without replacement, and $a\times {b \choose 2} +{14\choose 2 }$ to choose 2 ball with the same color.

Do some simplification I got a equation $\frac{n\times(b-1)}{n\times(n-1)}=\frac{n\times(b-1)+14\times 13}{(n+14)\times (n+13)}=\frac{14\times 13}{28n+14\times 13}=\frac{13}{2n+13}$ ,solve it and I got $n=\frac{13b}{15-2b}$ , then if $b=1,5,6,7$ then $n=1,13,26,91$ ,while $b$ is a divisors of $n$ and $n=1$ is nonsence for this problem, so the answer is 91

At the beginning we have m colours and n balls per colour in the urn, so m*n balls in total. I will put $m*n$ = $x$

The probability of drawing two balls, without replacement, of the same colour is $\frac {m*{n \choose 2} } {{mn \choose 2}}$ = $\frac{m*n*(n-1)}{m*n*(mn-1)}$ = $\frac {n-1}{x-1}$

Now we put in the urn 14 balls of another colour, different from the others, and recalculate the probability.

$\frac {m*{n \choose 2} + {14 \choose 2} }{ {mn+14 \choose 2} }$ = $\frac {x*(n-1) + 14*13} {(x+13)*(x+14)}$

These quantities are equals and, at the end, I obtain $x$ = $\frac{13n}{15-2n} \Rightarrow$ $m$ = $\frac{13}{15-2n}$

m and n are positive integers so there are two possibilities: n=1 m=1 (not accetable) and n=7 m=13

So the solution is $7*13$ = $91$

Suppose there are $mn$ balls of $n$ colors ( $m$ balls of each color). Then initially the probability of picking 2 balls of same color is given by $\dfrac{m-1}{mn-1}$ . Here you can pick any color ball the first time around, then pick the same color ball as the first ball out of the $mn-1$ balls remaining. For any color, $m-1$ of its kind would be in the urn.

Now when you have added 14 new balls to the urn, either you can pick out of the original colors or out of the new color. If the ball is of original color, first ball comes with probability $\dfrac{mn}{mn+14}$ and the same color second comes with $\dfrac{m-1}{mn+13}$ . If the ball is of the new color, first ball comes with probability $\dfrac{14}{mn+14}$ and the same color second comes with $\dfrac{13}{mn+13}$ . Totally probability for both colors is $\dfrac{mn(m-1)+14\times 13}{\left( mn+14 \right)\left( mn+13 \right)}$ . Both are given to be equal so

\begin{equation} \frac{m-1}{mn-1}=\dfrac{mn(m-1)+14\times 13}{\left( mn+14 \right)\left( mn+13 \right)} \end{equation}

Simplifying

\begin{equation} \left( m-1 \right)\left( mn+14 \right)\left( mn+13 \right)=\left( mn-1 \right)\left( mn(m-1)+182 \right) \end{equation}

\begin{equation} \left( m-1 \right)\left( {{m}^{2}}{{n}^{2}}+27mn+182 \right)=\left( mn-1 \right)\left( {{m}^{2}}n-mn+182 \right) \end{equation}

\begin{equation} 2mn+13=15n \end{equation}

\begin{equation} n=\frac{13}{15-2m} \end{equation}

Since $n$ is a natural number, $15-2m = 1 \text{or} =13$ . Only the first choice makes sense in the context as the latter gives $m=1, n=1$ . So

\begin{equation} m=7,\,\ n=13,mn=91 \end{equation}

Let c is number of colour initially And n is the number of balls of each colour Then p(drawing two balls of same colour)=c n (n-1)/[(nc) (nc-1)] When14 balls are added then total balls are nc+14 No of colour is c+1 Again p(drawing two balls of same colour) ={c n (n-1)+14 13}/{(nc+14)*(nc+13)} Equating both probability We get c=13/(15-2n) or n=0 As we know n is not 0 So 13 is a prime thus 15-2n is 1 or 13 So n=1 or n=7 But n should be greater than or equal to 2 to take out two balls os same colour So at n=7 c=13 So total balls =nc=91 answer

Let c be the number of colors Let n be the number of balls per color Thus cn is the original total number of balls

Originally, the probability of getting a pair of the same color would be: p = {n \choose 2}*c/{cn \choose 2}

When simplified becomes, (n-1)/(cn-1)

After adding the 14 balls, the new probability will be: p = (c{n \choose 2}+{14 \choose 2})/{(cn+14) \choose 2} When simplified becomes, (cn^2-cn+182)/c^2n^2 +27cn +182

Since the two probabilities are equal, equate them. After cross multiplication and some cancellation: 183cn -cn^2 = 27cn^2 +182n -27cn

183c - cn = 27cn +182 - 27c

210 c =28cn +182

2c(105 - 14n) = 182

c(105 - 4n) = 91 = 1x91 = 13x7 We know that c and n are both positive integers, the only pairs that work are n = 1 c = 91 and n = 7 c = 13 The former won't satisfy the conditions, but the latter will. Thus, cn = 91

Let there be $N$ balls and $x$ colours initially.

Therefore, there are $\left( \dfrac{N}{x} \right)$ balls of each of the $x$ colours. The probability of picking up two balls of the same colour is

$\Rightarrow 1 \times \dfrac{\left( \dfrac{N}{x} \right) - 1}{N-1}$

(The first ball that is picked could be of any colour, we need to only ensure that the second one is the same as the first colour)

Now, after the addition of the next unique colour, there are a total of $N + 14$ balls and $x + 1$ colours. Now, the probability of picking two colours is

$P(\text{two of the same colour}) = P(\text{both are the new colour}) + P(\text{the old colours})$

We can calculate this as follows

$\Rightarrow P(\text{both are the new colour}) = \dfrac{14}{N+14} \times \dfrac{13}{N+13}$

$\Rightarrow P(\text{the old colours}) = \dfrac{N}{N+14} \times \dfrac{\left( \dfrac{N}{x} \right) - 1}{N+13}$

Thus, the total probability after the addition of the new balls is

$\Rightarrow \left( \dfrac{14}{N+14} \times \dfrac{13}{N+13} \right) + \left( \dfrac{N}{N+14} \times \dfrac{\left( \dfrac{N}{x} \right) - 1}{N+13} \right)$

This is equal to the initial probability as described above.

Thus, we get

$\Rightarrow \left( \dfrac{14}{N+14} \times \dfrac{13}{N+13} \right) + \left( \dfrac{N}{N+14} \times \dfrac{\left( \dfrac{N}{x} \right) - 1}{N+13} \right) = \dfrac{\left( \dfrac{N}{x} \right) - 1}{N-1}$

When we solve this (I used wolfram alpha ), we get the solution as

$N = 15m + 1$ and $x = 2m + 1$ for $m \geq 1$ .

Also notice that we need to find an $m$ such that $\dfrac{N}{x}$ is an integer (meaning that $\dfrac{15m + 1}{2m + 1}$ should be an integer), since there are definitely an integer number of balls of every colour initially.

We can find the lowest (and perhaps the only) $m$ that satisfies this condition using a computer program, like the one below

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def is_int(x):
    return x%1.0 == 0

x = 1

while True:
    val = (15.0*x + 1)/(2.0*x + 1)
    if is_int(val):
        print (15.0*x + 1)
        break
    else: x += 1

#output 91.0

Thus, our final answer is $\boxed{91}$ . We could probably find a better way to find the lowest $N$ that satisfies the conditions using a more clever method.