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Let's go through the questions based on the images you have uploaded.

### From the first image:
1. **Sir Knock's desk drawer has 3 blue pens and 3 red pens**.
   - **3.2.1**: Write down the total number of pens in the desk drawer.
     - **Solution**: Total number of pens = 3 blue pens + 3 red pens = **6 pens**.
   
   - **3.2.2**: If a pen is picked from the desk drawer, write down the probability that it is red.
     - **Solution**: The probability of picking a red pen is given by the ratio of red pens to total pens:
       \[
       P(\text{Red}) = \frac{\text{Number of red pens}}{\text{Total number of pens}} = \frac{3}{6} = \frac{1}{2}
       \]
       So, the probability is **\( \frac{1}{2} \)**.

### From the second image:
1. **4.1**: Two events \( A \) and \( B \) are given with probabilities \( P(A) = \frac{2}{3} \) and \( P(B) = \frac{1}{5} \).
   - **4.1.1**: Calculate \( P(A \text{ and } B) \) if the events are mutually exclusive (disjoint).
     - **Solution**: If two events are mutually exclusive, they cannot happen at the same time. Therefore:
       \[
       P(A \text{ and } B) = 0
       \]

   - **4.1.2**: Calculate \( P(A \cup B) \), where \( A \cup B \) is the probability of either \( A \) or \( B \) occurring.
     - **Solution**: The formula for \( P(A \cup B) \) when events are mutually exclusive is:
       \[
       P(A \cup B) = P(A) + P(B)
       \]
       Substituting the given probabilities:
       \[
       P(A \cup B) = \frac{2}{3} + \frac{1}{5} = \frac{10}{15} + \frac{3}{15} = \frac{13}{15}
       \]
       So, \( P(A \cup B) = \frac{13}{15} \).

2. **4.2**: Events \( A \) and \( B \) are **not** mutually exclusive with \( P(A \cup B) = 0.84 \), \( P(A) = 0.67 \), and \( P(B) = 0.58 \).
   - **Solution**: To calculate \( P(A \cap B) \), we use the formula:
     \[
     P(A \cup B) = P(A) + P(B) - P(A \cap B)
     \]
     Rearranging the formula to solve for \( P(A \cap B) \):
     \[
     P(A \cap B) = P(A) + P(B) - P(A \cup B)
     \]
     Substituting the values:
     \[
     P(A \cap B) = 0.67 + 0.58 - 0.84 = 1.25 - 0.84 = 0.41
     \]
     So, \( P(A \cap B) = 0.41 \).

3. **4.3**: Consider three events where:
   - \( P(A) = 0.30 \)
   - \( P(B) = 0.55 \)
   - \( P(A \cup B) = 0.74 \)
   - **Solution**: Again, use the formula for the union of two events:
     \[
     P(A \cup B) = P(A) + P(B) - P(A \cap B)
     \]
     Rearranging to solve for \( P(A \cap B) \):
     \[
     P(A \cap B) = P(A) + P(B) - P(A \cup B)
     \]
     Substituting the given values:
     \[
     P(A \cap B) = 0.30 + 0.55 - 0.74 = 0.85 - 0.74 = 0.11
     \]
     So, \( P(A \cap B) = 0.11 \).

### Summary of answers:
1. **Total number of pens** = 6 pens.
2. **Probability of picking a red pen** = \( \frac{1}{2} \).
3. **4.1.1**: \( P(A \cap B) = 0 \) (mutually exclusive).
4. **4.1.2**: \( P(A \cup B) = \frac{13}{15} \).
5. **4.2**: \( P(A \cap B) = 0.41 \).
6. **4.3**: \( P(A \cap B) = 0.11 \).

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Let me know if you'd like further clarifications or more detailed explanations!

#### Here are 5 relative questions:
1. How do we determine if two events are mutually exclusive?
2. What is the general formula for calculating the probability of the union of two events?
3. How do we calculate the probability of the intersection of two independent events?
4. What changes in the calculation when events are not mutually exclusive?
5. Can probabilities be greater than 1? Why or why not?

**Tip**: For mutually exclusive events, the probability of both events happening at the same time is always zero!

Sir Knock's desk drawer has 3 blue pens and 3 red pens. Calculate the total number of pens and the probability of picking a red pen. Also, solve probability questions related to events A and B, including mutually exclusive and non-mutually exclusive events.

Solve probability questions involving 3 blue and 3 red pens, including calculating the probability of picking a red pen. Additionally, explore advanced concepts with events A and B, focusing on mutually exclusive and non-mutually exclusive events.

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