3.	At a fair, each person can spin two wheels of chance. The first wheel has the numbers 1, 2, and 3. The second wheel has the letters A and B. 
(a)	Explain why this is an independent event or a dependent event and list all the possible combinations of spinning both wheels.
(b)	If you spin both wheels, what is the probability that you get either a 1 or an A? Explain.


### Part (a): Independent or Dependent Event & Possible Combinations

The event of spinning the two wheels is **independent**. This is because the outcome of spinning the first wheel (which has numbers) does not affect the outcome of spinning the second wheel (which has letters). The result of each spin is determined independently of the other.

#### List of Possible Combinations:
Since the first wheel has 3 possible outcomes (1, 2, 3) and the second wheel has 2 possible outcomes (A, B), the total number of combinations is:

\[
3 \times 2 = 6 \text{ combinations}
\]

The possible combinations are:

- (1, A)
- (1, B)
- (2, A)
- (2, B)
- (3, A)
- (3, B)

### Part (b): Probability of Getting Either a 1 or an A

To calculate the probability of getting either a 1 or an A when spinning both wheels, we first need to figure out how many outcomes meet this condition. 

1. **Probability of getting a 1:**  
   The first wheel can land on 1. The outcomes that include a 1 are:
   - (1, A)
   - (1, B)  
   So there are 2 outcomes with a 1.

2. **Probability of getting an A:**  
   The second wheel can land on A. The outcomes that include an A are:
   - (1, A)
   - (2, A)
   - (3, A)  
   So there are 3 outcomes with an A.

3. **Avoiding Double Counting (overlap):**  
   We already counted the outcome (1, A) twice (once for getting a 1 and once for getting an A), so we subtract it once to avoid double counting.

Thus, the total number of favorable outcomes is:
\[
2 \text{ (for 1)} + 3 \text{ (for A)} - 1 \text{ (overlap)} = 4 \text{ favorable outcomes}
\]

Since there are 6 total possible outcomes, the probability of getting either a 1 or an A is:
\[
P(\text{1 or A}) = \frac{4}{6} = \frac{2}{3}
\]

### Summary:
- (a) This is an independent event, and the possible combinations are (1, A), (1, B), (2, A), (2, B), (3, A), (3, B).
- (b) The probability of getting either a 1 or an A is \( \frac{2}{3} \).

Would you like more details or have any questions?

Here are 5 related questions to expand on this:
1. What is the probability of getting exactly a 2 and a B?
2. What is the probability of getting neither a 1 nor an A?
3. If a third wheel is added with outcomes X, Y, Z, how many possible combinations will there be?
4. How does the concept of independent and dependent events apply to real-life situations?
5. What would happen to the probability if there were more numbers or letters on the wheels?

**Tip**: When calculating the probability of "either A or B," always remember to account for any overlap between A and B to avoid double counting!

At a fair, each person can spin two wheels of chance. The first wheel has the numbers 1, 2, and 3. The second wheel has the letters A and B. (a) Explain why this is an independent event or a dependent event and list all the possible combinations of spinning both wheels. (b) If you spin both wheels, what is the probability that you get either a 1 or an A? Explain.

Explore how to calculate the probability of getting either a 1 or an A when spinning two independent wheels at a fair. The first wheel has numbers 1, 2, 3, and the second wheel has letters A and B. This problem focuses on probability, independent events, and combinatorics. Suitable for Grades 6-8.

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