Suppose a bag contains 7 blue chips and 3 gold chips. What the probability randomly choosing a gold chip, not replacing it, and then randomly choosing another gold chip?

$\frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$

Explanation:

There are 10 chips in total.

On the first draw, there are 3 gold chips and so the probability of choosing a gold chip is:

${P}_{1} = \frac{3}{10}$

We don't replace the chip, and so now there are 2 gold chips and 9 chips total. We draw again for a gold chip:

${P}_{2} = \frac{2}{9}$

The probability then of drawing 2 gold chips in a row is:

$\frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$