Q:

If five numbered balls are taken at random, find probability that the second largest number is 8

An urn contains ten balls numbered $1$ through $10$. Five balls are drawn at random and without replacement. Find probability that the second largest number drawn will be $8$.

The root of my confusion lies in whether the balls are drawn one at a time, or scooped up at once. 

A:

Addressing your comment as to whether the balls are drawn one at a time, or together: it doesn't matter.
If you draw the balls one at a time, you would probably think of choosing five numbers from ten, including the number $8$, with order being important.  To count the number of favourable draws,

choose a place for the $8$. . . . . . $5$ ways
choose a digit greater than $8$ (so that $8$ is the second largest). . . . . . $2$ ways
choose a place for this digit. . . . . . $4$ ways
choose three digits below $8$, with order important. . . . . . $7\times6\times5$ ways.

To count the total number of draws:

choose five digits, order important. . . . . . $10\times9\times8\times7\times6$ ways.

The probability is
$$\frac{5\times2\times4\times7\times6\times5}{10\times9\times8\times7\times6}=\frac{5}{18}\ .$$
If you draw them all together you are probably thinking of order as being irrelevant.  The number of favourable draws is $2C(7,3)$, the total number of draws is $C(10,5)$, and the probability is
$$\frac{2C(7,3)}{C(10,5)}=\frac{2\times7\times6\times5}{3\times2\times1}\frac{5\times4\times3\times2\times1}{10\times9\times8\times7\times6}=\frac{5}{18}\ .$$

