To solve this problem, we calculate the probability step by step.

Step 1: Determine the total number of cookies in the bag.

The bag contains:

• $9$ chocolate chip cookies,
• $8$ peanut butter cookies,
• $6$ sugar cookies,
• $9$ oatmeal cookies.

Thus, the total number of cookies is: $9 + 8 + 6 + 9 = 32$

Step 2: Calculate the probability of selecting a chocolate chip cookie first.

The probability of selecting a chocolate chip cookie is: $\frac{\text{Number of chocolate chip cookies}}{\text{Total number of cookies}} = \frac{9}{32}.$

Step 3: Calculate the probability of selecting an oatmeal cookie after the chocolate chip cookie is eaten.

After eating one chocolate chip cookie, the total number of cookies decreases to: $32 - 1 = 31.$ The number of oatmeal cookies remains $9$. The probability of selecting an oatmeal cookie is now: $\frac{\text{Number of oatmeal cookies}}{\text{New total number of cookies}} = \frac{9}{31}.$

Step 4: Multiply the probabilities.

Since the two events are independent, we multiply the probabilities: $\text{Overall probability} = \frac{9}{32} \cdot \frac{9}{31} = \frac{81}{992}.$

Final Answer:

The probability that Fatoumata randomly selects a chocolate chip cookie, eats it, then selects an oatmeal cookie is: $\boxed{\frac{81}{992}}$

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Would you like me to explain any part of this solution further?

Related Questions:

1. What is the probability of selecting two oatmeal cookies in a row?
2. How would the probability change if Fatoumata does not eat the first cookie?
3. What is the probability of selecting one peanut butter and one sugar cookie in sequence?
4. What is the total probability of selecting any type of cookie twice in a row?
5. How does the probability change if the number of chocolate chip cookies is increased to 10?

Tip:

When calculating probabilities in sequence, always adjust the total number of possibilities after each selection if the item is not replaced.