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How many students were observed?
*P*(no instrument)
*P*(no hat)
*P*(no instrument and no hat)
*P*(no instrument∣no hat)
*P*(no hat∣no instrument)
Are the events carrying a musical instrument and wearing a hat independent?

To answer the questions, we'll start by calculating the total number of students observed and then use the given data to find the required probabilities.

**Step 1: Total Number of Students Observed**

The data provided is as follows:

|                         | Hat | No Hat | Total |
|-------------------------|-----|--------|-------|
| Musical instrument      | 112 | 248    | 360   |
| No musical instrument   | 68  | 172    | 240   |
| Total                   | 180 | 420    | 600   |

So, the total number of students observed is 600.

**Step 2: Calculating Probabilities**

1. **\( P(\text{No instrument}) \)**

\[ P(\text{No instrument}) = \frac{\text{Number of students with no musical instrument}}{\text{Total number of students}} \]
\[ P(\text{No instrument}) = \frac{240}{600} = 0.4 \]

2. **\( P(\text{No hat}) \)**

\[ P(\text{No hat}) = \frac{\text{Number of students with no hat}}{\text{Total number of students}} \]
\[ P(\text{No hat}) = \frac{420}{600} = 0.7 \]

3. **\( P(\text{No instrument and no hat}) \)**

\[ P(\text{No instrument and no hat}) = \frac{\text{Number of students with no musical instrument and no hat}}{\text{Total number of students}} \]
\[ P(\text{No instrument and no hat}) = \frac{172}{600} \approx 0.287 \]

4. **\( P(\text{No instrument} \mid \text{No hat}) \)**

\[ P(\text{No instrument} \mid \text{No hat}) = \frac{P(\text{No instrument and no hat})}{P(\text{No hat})} \]
\[ P(\text{No instrument} \mid \text{No hat}) = \frac{0.287}{0.7} \approx 0.41 \]

5. **\( P(\text{No hat} \mid \text{No instrument}) \)**

\[ P(\text{No hat} \mid \text{No instrument}) = \frac{P(\text{No instrument and no hat})}{P(\text{No instrument})} \]
\[ P(\text{No hat} \mid \text{No instrument}) = \frac{0.287}{0.4} \approx 0.717 \]

**Step 3: Checking Independence**

Two events \( A \) and \( B \) are independent if and only if \( P(A \cap B) = P(A)P(B) \).

For this problem:

\[ P(\text{Instrument and Hat}) = \frac{112}{600} \approx 0.187 \]
\[ P(\text{Instrument}) = \frac{360}{600} = 0.6 \]
\[ P(\text{Hat}) = \frac{180}{600} = 0.3 \]

If \( P(\text{Instrument}) \) and \( P(\text{Hat}) \) are independent, then:

\[ P(\text{Instrument and Hat}) = P(\text{Instrument}) \times P(\text{Hat}) \]
\[ 0.187 \neq 0.6 \times 0.3 \]
\[ 0.187 \neq 0.18 \]

Since \( P(\text{Instrument and Hat}) \neq P(\text{Instrument}) \times P(\text{Hat}) \), the events are not independent.

**Summary:**

1. Total students observed: 600
2. \( P(\text{No instrument}) = 0.4 \)
3. \( P(\text{No hat}) = 0.7 \)
4. \( P(\text{No instrument and no hat}) \approx 0.287 \)
5. \( P(\text{No instrument} \mid \text{No hat}) \approx 0.41 \)
6. \( P(\text{No hat} \mid \text{No instrument}) \approx 0.717 \)
7. The events of carrying a musical instrument and wearing a hat are not independent.

Would you like more details or have any questions?

Here are 5 additional questions you might ask:

1. How do we determine if two events are mutually exclusive?
2. Can you explain the difference between independent and mutually exclusive events with examples?
3. How would the calculations change if we had more detailed categories for hats and instruments?
4. What is the significance of conditional probability in real-world applications?
5. How can we use Venn diagrams to visualize these probabilities and relationships?

**Tip:** When working with probabilities, carefully distinguish between the total probability of an event and the conditional probability to avoid confusion.

Explore the detailed solution to calculating probabilities based on student observations with and without musical instruments and hats. Includes conditional probability calculations and determination of event independence.

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