Question

Tabitha and her mother went to the pet store. Her mother told her she could either choose a cat, dog, or fish, and either get a toy for the pet or a book about the pet.

What is the probability that she chooses a four-legged pet and a toy for it?


A.


B.


C.


D.

Answer 1

To find the probability that Tabitha chooses a four-legged pet (which can only be a cat or a dog) and chooses a toy for it, we need to analyze the options available to her.

1. **Types of Pets**:
- Cat (4-legged)
- Dog (4-legged)
- Fish (not 4-legged)

Hence, there are 2 options for four-legged pets (cat and dog) and 1 option for a non-four-legged pet (fish).

2. **Choices of Items**:
- For the chosen pet, she can choose either a toy or a book.
- Therefore, for each pet type, she has 2 options:
- Toy
- Book

So, let's summarize:
- If she chooses a cat:
- Cat + Toy
- Cat + Book
- If she chooses a dog:
- Dog + Toy
- Dog + Book
- If she chooses a fish:
- Fish + Toy
- Fish + Book

From this, we can see the total combinations are:
- Cat + Toy
- Cat + Book
- Dog + Toy
- Dog + Book
- Fish + Toy
- Fish + Book

Thus, there are a total of \( 3 \text{ (pets)} \times 2 \text{ (items per pet)} = 6 \text{ total combinations} \).

Now, we want to find only the combinations where she chooses a four-legged pet and a toy:
1. Cat + Toy
2. Dog + Toy

There are 2 combinations that fit the criteria (four-legged pet + toy).

Now we can calculate the probability:
- Total favorable outcomes = 2 (Cat + Toy and Dog + Toy)
- Total possible outcomes = 6

Thus, the probability \( P \) is given by:
\[
P(\text{four-legged pet and toy}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{2}{6} = \frac{1}{3}
\]

Therefore, the probability that Tabitha chooses a four-legged pet and a toy for it is \(\frac{1}{3}\).

If options A, B, C, and D correspond to specific numerical values or other formats, please provide them, and I can help you identify the correct one.