To solve this problem, we need to find the probability that Kori selects 2 peanut butter cookies from the bag.

Step 1: Calculate the total number of cookies.

• Chocolate chip cookies: 8
• Peanut butter cookies: 4
• Sugar cookies: 8
• Oatmeal cookies: 9

The total number of cookies in the bag is: $8 + 4 + 8 + 9 = 29 \text{ cookies}$

Step 2: Calculate the number of ways to choose 2 peanut butter cookies from 4.

The number of ways to choose 2 peanut butter cookies from 4 is given by the combination formula $\binom{n}{k}$, where $n$ is the total number of peanut butter cookies, and $k$ is the number of cookies we want to choose: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$

Step 3: Calculate the total number of ways to choose any 2 cookies from 29.

The number of ways to choose 2 cookies from 29 is: $\binom{29}{2} = \frac{29!}{2!(29-2)!} = \frac{29 \times 28}{2 \times 1} = 406$

Step 4: Calculate the probability.

The probability that Kori selects 2 peanut butter cookies is the ratio of the favorable outcomes to the total outcomes: $P(\text{2 peanut butter cookies}) = \frac{\binom{4}{2}}{\binom{29}{2}} = \frac{6}{406} \approx 0.0148$

So, the probability is approximately 0.0148.

Would you like more details or have any questions? Here are some related questions:

1. What is the probability of selecting at least one peanut butter cookie?
2. What is the probability of selecting two chocolate chip cookies?
3. How would the probability change if Kori picks 3 cookies instead of 2?
4. What is the probability of selecting 1 sugar cookie and 1 oatmeal cookie?
5. How many different pairs of cookies can Kori select from the bag?

Tip: When calculating combinations, always remember that order does not matter, which is why factorials are used to eliminate permutations.