To solve this problem, we need to determine the probability of two events happening in sequence:

1. Damien selects a chocolate chip cookie from the bag and eats it.
2. He then selects a sugar cookie from the remaining cookies.

Step 1: Find the total number of cookies.

Damien’s bag contains the following cookies:

• 8 chocolate chip cookies
• 4 peanut butter cookies
• 8 sugar cookies
• 8 oatmeal cookies

The total number of cookies is:

$8 + 4 + 8 + 8 = 28$

Step 2: Probability of selecting a chocolate chip cookie.

Initially, there are 8 chocolate chip cookies out of the 28 total cookies. The probability of selecting a chocolate chip cookie is:

$\frac{8}{28} = \frac{2}{7}$

Step 3: Probability of selecting a sugar cookie after eating the chocolate chip cookie.

After eating the chocolate chip cookie, there are now 27 cookies left in the bag (one fewer). The number of sugar cookies remains 8, so the probability of selecting a sugar cookie is:

$\frac{8}{27}$

Step 4: Find the combined probability.

To find the probability of both events happening in sequence, we multiply the probabilities of each event:

$\frac{2}{7} \times \frac{8}{27} = \frac{16}{189}$

Thus, the probability that Damien selects a chocolate chip cookie first and then a sugar cookie is:

$\boxed{\frac{16}{189}}$

Would you like further details on any of the steps? Here are five related questions you might consider:

1. What is the probability Damien selects two chocolate chip cookies in a row?
2. What is the probability he selects an oatmeal cookie first, then a peanut butter cookie?
3. If Damien eats two cookies randomly, what is the probability that neither is a sugar cookie?
4. What is the probability he selects one peanut butter cookie and one sugar cookie?
5. How would the probability change if there were 10 chocolate chip cookies instead?

Tip: Always reduce fractions by finding the greatest common divisor (GCD) to express your answer in the simplest form.