Bizarre Bag of Balls

First, we must find the probability of the sum is $n+1$ . That sum can be found with the following ordered ball values taken from the bags: $\{(1,n),(2,n-1),(3,n-2),...,(n,1)\}$

One can see this set has exaclty $n$ elements, and the the number of possible ordered pairs is $n^2$ , so our desired probabillity is $\frac{n}{n^2} = \frac{1}{n}$ .

Using the same method for a sum value of $j$ , we have the pairs: $\{(1,j-1),(2,j-2),...,(j-1,1)\}$ Again, this set has $j-1$ elements and the probability of sum $j$ is $\frac{j-1}{n^2}$ .

Our desired sum could be any integer $j \leq k$ , so we must sum the probabilities for each $j \in \{2,3,...,k\}$ . We have our final probability given by: $\sum_{j=1}^k\frac{j-1}{n^2} = \frac{(j-1)j}{n^2}$ and here we used the closed formula for arithmetic progression sum, or more simply, the well know formula to sum the first few natural numbers.

From now on, we do some algebra:

$\frac{(j-1)j}{n^2} = \frac{1}{n} \Rightarrow (k-1)k = 2n$

If we notice that $(k-1)k$ is always even, $n$ will be integer for every $k$ . We just need to find out the maximum $k$ , and this is quite simple.

$n\leq 500 \Rightarrow 2n \leq 1000 \Rightarrow k(k-1) \leq 1000$ $(k-1)^2 \leq k(k-1) \leq 1000 \Rightarrow (k-1)^2 \leq 1000$ $k-1 \leq \sqrt(1000) \Rightarrow k-1 \leq 31 \Rightarrow k\leq 32$

Remembering that every $k\geq 2$ will give us a valid $n$ , we find $32-2+1 = 31$ possible values for $n$ .

Of the n^2 possible ordered pairs, only n pairs add up to n+1. This is because given any first ball x, the second must be (n + 1 - x), all of which values are between 1 and n.

Now, count pairs of balls that add up to $\leq$ k; call this number pairs(k). For the condition to be true, pairs(k) must be equal to n.

pairs(k) is the number of ordered pairs of positive integers (a, b) such that a + b $\leq$ k. For each a < k, b can have values from 1 to (k - a). So pairs(k) is the sum of the (k-1) consecutive integers from 1 to (k-1), or the (k - 1)th triangular number. Since n has to be equal to pairs(k), n has to be a triangular number. There are 31 triangular numbers between 1 and 500, so the answer is 31 possible values of n.

Specifically, for any integer k > 1, the probability is equal when $n = \frac {(k-1)k} {2}$

First, observe that the probability the sum is $n+1$ is $\frac{1}{n}$ . Regardless of what numbered ball we draw out of the first bag, there is exactly one ball from the second bag that will give a sum of $n+1$ .

Next, we calculate the probability that the sum is less than or equal to $k$ . We assume that $k\le n+1$ . (If $k>n+1$ , the probability is always $1$ .) If ball $i$ , $1\le i\le k-1$ , is chosen from the first bag, there are $k-i$ balls in the second bag that will give us a sum less than or equal to $k$ (the choices range from $1$ to $k-i$ inclusive). Hence, the probability that the sum is less that $k$ is $\displaystyle\sum^{k-1}_{i=1}\left(\frac{1}{n}*\frac{k-i}{n}\right)=\frac{k(k-1)}{2n^2}$ .

We want the two probabilities to be equal, so $\frac{k(k-1)}{2n^2}=\frac{1}{n}\implies\frac{k(k-1)}{2}=n$ . Since $1\le n\le 500$ , it follows that $2\le k\le 32$ , so there are $\boxed{31}$ values of $n$ .

For any number chosen for the first ball, there is exactly one ball from the other bag which makes the sum n+1, n+1-whatever you chose the first time. That means the probability of choosing the number which makes the sum n+1 is 1/n because regardless the value of the first ball, there is exactly one second ball out the n balls in the second bag to choose in order to make the sum n+1. Now, the probability the sum is less than or equal to k is the probability of choosing some ball i less than or equal to k-1 and then choosing any ball less than or equal to k-i in the next bag. This works out to (k,2) and there are n^2 ways to choose the two balls so the probability is (k,2)/(n^2). we want 1/n=(k,2)/(n^2) which simplifies to n=(k,2). so we must find numbers expressible in the form (k,2). The largest of these is 31 because (31,2) is 496, the largest number expressible in form (k,2) less than 500 because n must be less than or equal to 500. therefore, all the numbers 1...31 work, meaning there are 31 solutions.

it is easy to know that the probability of getting a sum of n+1 is equal to 1/n.

we want to find k such that the probability of getting a sum of k is also equal to 1/n. we know that the smallest possible sum is 2 and the highest possible sum is 2n. the probability of getting a sum x (x is an integer) is (x-1)/(n^2) for x<(n+1) and (2n-x)/(n^2) for x>(n+1). let y be equal to summation of all x from 2 until k. therefore, the probability of getting a sum of less than or equal to k is equal y/(n^2). since y/(n^2) must be equal to 1/n, y=n.

from here, we can conclude that 1<n<500 and there exist a number z such that n = summation from 1 to z. the highest possible value of z that would satisfy these conditions is 31. thus, n must be 31.

Let x and y be number on balls drawn. Total ways to draw balls=n^2. Favourable ways in which x+y=n+1 are n. Probability of getting sum n+1 is 1/n. For sum to be less than or equal to k x + y<=k. Total ways to select x and y satisfying above equation=k(k-1)/2. Probability of getting sum less than or equal to k=k(k-1)/2n^2. Since probability of getting sum less than or equal to k=probability of getting sum n+1. Thus, k(k-1)/2n^2=1/n or k(k-1)/2= Since, n<=500 k(k-1)<=1000 k<=32 For n>0, k>=2. Thus, each value of k lying between 2 and 32 inclusive gives a value of n satisfying above conditions. Thus there are 31 values of n satisfying required conditions.

The probability of getting a sum of $n+1$ when two balls are selected is $\frac{n}{n^2}$ . For $2 \leq m \leq n+1$ , the probability of getting a sum of $m$ is $\frac{m-1}{n^2}$ . So for any $k$ , the probability that the sum is $\leq k$ is $\frac{\sum_{i=2}^{k}(i-1)}{n^2} = \frac{\sum_{i=1}^{k-1} i}{n^2} = \frac{(k)(k-1)}{2n^2}$ .

If we have $\frac{n}{n^2} = \frac{(k)(k-1)}{2n^2}$ then $n = \frac{(k)(k-1)}{2}$ , so as long as $n$ is a triangular number, such a $k$ will exist. For $k=1$ , we get $n=0$ , which is out of range. For $k =2$ , we get $n=1$ . On the other hand, for $k = 32$ , we get $n = 496$ . For $k=33$ , we get $n = 528$ , which is out of range. Thus, there are $32 - 2 + 1 = 31$ values of $n$ that will work.

Let $A$ and $B$ be the set of $n$ balls in each bag. Consider the set $A\times B$ of all possible pairs of balls, one from each bag.

Now, from the total of pairs, exactly $n$ of them sum up $n+1$ , and, we may consider $n>2$ by now.

For the sum of the balls to be equal to $k<n+1$ , we have in total $k$ possible pair of balls, and thus, since $\frac{n}{n^{2}}=\frac{1}{n}<\frac{1}{2}$ , we may just consider the sums less than $n+1$ .

We have then that for the sum to be at most $k$ , we have

$1+2+\cdots+k=\frac{k\left(k+1\right)}{2}$ forms.

Then, we must have

$\frac{k\left(k+1\right)}{2}=n$

and thus $n$ is a triangular number, and we have exactly $30$ triangular numbers less than $500$ and greater than $1$ . Now if $n=1, 2$ , we have that for $n=1$ , $k=2$ works, and $n=2$ does not work, so there is a total of $31$ possible numbers.

Let $$ABC$$ be the required triangle.Let us take the two points A and B randomly subtending $$\angle \theta$$ at centre O.Then, probability of acute angled triangle is $$P (\theta)=\frac {\theta}{2 \pi}$$ because the point C must be within the portion of arc between the two points diametrically opposite A and B(say $$A^'$$ and $$B^'$$ respectively)Again probability of the angle between A and B lying between $$'\theta' and 'd+d \theta'$$ is $$\frac {d \times \theta}{\pi}$$ Thus total probability $$P=\displaystyle \int_{\theta=0}^\pi {\frac {\theta}{2 \pi}} \times {\frac {d \theta}{\pi}}$$ $$=\frac {1}{4}$$