Drawing balls without replacement

Since every product is equally likely, the expected value of $xy$ is the average of all possible products. Also, let's assume that the order of $x$ and $y$ matters. We compute the sum of the products by strategically over-counting and then subtracting.

First, we compute the sum of the products, assuming that $x$ and $y$ can be equal. Then notice the expression $(1+2+\ldots+7)(1+2+\ldots+7),$ when expanded, includes every single product. Therefore, the sum of these products is $(1+2+\ldots+7)(1+2+\ldots+7) = 28 \times 28 = 784.$

Then, we subtract the products in which $x$ and $y$ are equal. These products have sum $1^2 + 2^2 + \ldots + 7^2 = 140,$ so subtracting gives $784 - 140 = 644$ as the final total.

There are $7 \times 6 = 42$ ways to choose $x$ and $y.$ Therefore, the average of the products is $\dfrac{644}{42} = \dfrac{46}{3}.$ The answer is $46 + 3 = \boxed{49}.$

The sum of all possible values of $xy$ is equal to $\begin{aligned}\sum_{x=1}^7 x\biggr[ \sum_{y=1}^7 y - x \biggr] &= \sum_{x=1}^7 x\biggr[\frac{7\times 8}{2} - x \biggr] \\ &= \sum_{x=1}^7 28x - x^2 \\ &= 28^2 - \dfrac{7\times 8\times 15}{6} \\ &= 644\end{aligned}$ Each outcome has probability equal to $\dfrac{1}{7\times 6} = \dfrac{1}{42}$ . Therefore, the expected value is $\dfrac{644}{42} = \dfrac{46}{3} = \dfrac{a}{b}$ , hence $a + b = \boxed{49}$

The sample space of possible choices is listed below. The first number in each ordered pair represents the first ball chosen and likewise for the second ball. $\begin{matrix} (1,2)&(2,1)&(3,1)&(4,1)&(5,1)&(6,1)&(7,1) \\ (1,3)&(2,3)&(3,2)&(4,2)&(5,2)&(6,2)&(7,2) \\ (1,4)&(2,4)&(3,4)&(4,3)&(5,3)&(6,3)&(7,3) \\ (1,5)&(2,5)&(3,5)&(4,5)&(5,4)&(6,4)&(7,4) \\ (1,6)&(2,6)&(3,6)&(4,6)&(5,6)&(6,5)&(7,5) \\ (1,7)&(2,7)&(3,7)&(4,7)&(5,7)&(6,7)&(7,6) \\ \end{matrix}$ Each of these pairs has probability $\frac{1}{42}$ of occurence. We are trying to find the sum of the products of the numbers in each pair, so that dividing this sum by 42 results in the desired probability. Rather than computing all of these products and then finding their sum, we can see that there exists symmetry in the pairs; the sum of the products of the numbers in the pairs of the $n^\text{th}$ column in the above sample space can be represented as $1n+2n+3n..+7n-n^2=n(1+2+3...+7-n)=n(28-n)$ . Our sum simplifies to $\displaystyle \sum_{n=1}^7 n(28-n)$ . $\begin{aligned} \sum_{n=1}^7 {n(28-n)}= \sum_{n=1}^7 {28n-n^2} \\ = \sum_{n=1}^7 28n- \sum_{n=1}^7 n^2 \\ = 28 \sum_{n=1}^7 n- \sum_{n=1}^7 n^2 \\ = 28\frac{7 \cdot 8}{2}- \frac{7 \cdot 8 \cdot 15}{6} \\ =644 \end{aligned}$ where the penultimate step is a result of using the sums of powers formulas (proven here in several ways).

The rest of the computation is trivial. $\frac{644}{42}=\frac{46}{3}$ , and the sum of the numerator and the denominator is $\boxed{49}$ .

Note that $1 + 2 + 3 + 4 + 5 + 6 + 7 = 7 \cdot 4 = 28$ , Also there are $6 \cdot 7 = 42$ possible draws. If we draw ball numbered $a$ , then $(28 - a) \cdot a$ is the sum of all possible values of $xy$ . Hence $$\frac{27 \cdot 1 + 26 \cdot 2 + 25 \cdot 3 + 24 \cdot 4 + 23 \cdot 5 + 22 \cdot 6 + 21 \cdot 7}{42} = \frac{46}{3}$$ Hence $$a + b = 46 + 3 = \boxed{49}$$

First, we calculate the total of all possible values of $x \cdot y$ . This can be written as:

$\sum_{x=1}^{7} \sum_{y=1}^{7} x \cdot y = 784$

As the balls are drawn out without replacement, this means that the two integers are distinct. Therefore, we must subtract:

$\sum_{x = 1}^7 x^2 = 140$

So the sum of all the possible values of $x \cdot y$ is $784 - 140 = 644$ . There are $7^2 - 7 = 42$ possible values of $x \cdot y$ , so to calculate the expected value, we do:

$\frac{644}{42} = \frac{46}{3}$

So the final answer is $46 + 3 = \fbox{49}$

There are $\dbinom {7}{2} = 21$ possibilities, but they are not necessarily all distinct. But each pair has probability $\dfrac {1}{21}$ . It doesn't matter whether they are distinct. We will not be overcounting or undercounting if we count each pair once. For example, $3 \cdot 4$ has probability $\dfrac {1}{21}$ and $2 \cdot 6$ has probability $\dfrac {1}{21}$ . The total probability is the sum of all the products (with repeats) divided by the total number of pairs, which is $\begin{aligned} & \dfrac {1 \cdot (2+3+4+5+6+7) + 2 \cdot (3 + 4 + 5 + 6 + 7) + \cdots + 6 \cdot 7}{21}\\ = & \dfrac {322}{21} \\ = & \dfrac {46}{3} \implies \boxed {49}. \end{aligned}$