The problem states:

• There are 7 puppies in total: 3 poodles, 3 terriers, and 1 retriever.
• Rebecka and Aaron each select one puppy at random, in that order, without replacement.
• We are asked to find the probability that Aaron selects the retriever, given that Rebecka selects a retriever.

Solution:

This is a conditional probability problem, where we are finding $P(\text{Aaron selects retriever} \mid \text{Rebecka selects retriever})$.

1. Total puppies: There are 7 puppies.
2. Case 1: Rebecka selects the retriever. The probability of Rebecka selecting the retriever is: $P(\text{Rebecka selects retriever}) = \frac{1}{7}$
3. Case 2: Aaron selects the retriever, given that Rebecka has already selected the retriever. Since the retriever has already been taken by Rebecka, Aaron cannot select the retriever. So, the probability that Aaron selects the retriever in this case is: $P(\text{Aaron selects retriever} \mid \text{Rebecka selects retriever}) = 0$

Thus, the probability is 0.

Would you like more details on this, or have any follow-up questions? Here are some related questions:

1. What is conditional probability, and how is it applied here?
2. How would the probability change if Rebecka had selected a different puppy?
3. How do probabilities change in scenarios of "selection without replacement"?
4. How would the problem change if there were more retrievers?
5. What if Aaron had selected first instead of Rebecka?

Tip: In probability problems involving "without replacement," always update the total options after each selection.