Iman’s favorite colors are 
\[\blue{\text{blue}}\] and 
\[\pink{\text{pink}}\].
She has 
\[\blue{\text{1 blue shirt}}\], 
\[\pink{\text{1 pink shirt}}\], 
\[\blue{\text{1 blue hat}}\], 
\[\blue{\text{1 blue belt}}\], 
\[\blue{\text{1 blue pair of pants}}\], and 
\[\pink{\text{1 pink pair of pants}}\].
Iman selects one of these garments at random. Let 
\[A\] be the event that she selects a pink garment and 
\[B\] be the event that she chooses a shirt.
What is 
\[P(A\text{ or }B)\], the probability that the garment Iman chooses is pink or a shirt? 


To solve for \( P(A \text{ or } B) \), the probability that Iman selects a pink garment or a shirt, we can use the following formula for the union of two events:

\[
P(A \text{ or } B) = P(A) + P(B) - P(A \cap B)
\]

### Step 1: List of Garments
Iman has the following garments:
- **1 blue shirt**
- **1 pink shirt**
- **1 blue hat**
- **1 blue belt**
- **1 blue pair of pants**
- **1 pink pair of pants**

Thus, there are a total of 6 garments.

### Step 2: Define the Events
- **Event \(A\):** Iman selects a **pink garment**. The pink garments are:
  - 1 pink shirt
  - 1 pink pair of pants
  
  So, \( A \) includes 2 garments. Therefore, the probability of selecting a pink garment is:
  \[
  P(A) = \frac{2}{6} = \frac{1}{3}
  \]

- **Event \(B\):** Iman selects a **shirt**. The shirts are:
  - 1 blue shirt
  - 1 pink shirt
  
  So, \( B \) includes 2 garments. Therefore, the probability of selecting a shirt is:
  \[
  P(B) = \frac{2}{6} = \frac{1}{3}
  \]

### Step 3: Intersection of \(A\) and \(B\) (i.e., \(A \cap B\))
- \( A \cap B \) represents the event that Iman selects a garment that is both pink and a shirt. This happens only if she selects the **pink shirt**, which is 1 garment.
  
  Therefore, the probability of selecting a garment that is both pink and a shirt is:
  \[
  P(A \cap B) = \frac{1}{6}
  \]

### Step 4: Apply the Formula
Now we can calculate \( P(A \text{ or } B) \) using the formula:

\[
P(A \text{ or } B) = P(A) + P(B) - P(A \cap B)
\]
\[
P(A \text{ or } B) = \frac{1}{3} + \frac{1}{3} - \frac{1}{6}
\]
\[
P(A \text{ or } B) = \frac{2}{3} - \frac{1}{6} = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
\]

### Conclusion:
The probability that Iman selects a pink garment or a shirt is \( \frac{1}{2} \).

Would you like more details on any part of this solution, or do you have any questions?

### Related Questions:
1. What is the probability that Iman selects a blue garment?
2. What is the probability that Iman selects a garment that is neither pink nor a shirt?
3. How does the probability change if another pink garment is added?
4. What is the probability that Iman selects a pink pair of pants or a blue shirt?
5. If Iman has 2 more blue shirts, how does \( P(B) \) change?

**Tip:** When calculating probabilities involving unions, always account for the intersection to avoid double-counting.

Iman’s favorite colors are blue and pink. She has 1 blue shirt, 1 pink shirt, 1 blue hat, 1 blue belt, 1 blue pair of pants, and 1 pink pair of pants. Iman selects one of these garments at random. Let A be the event that she selects a pink garment and B be the event that she chooses a shirt. What is P(A or B), the probability that the garment Iman chooses is pink or a shirt?

This problem calculates the probability that Iman selects a pink garment or a shirt from her wardrobe using the inclusion-exclusion principle. Learn how to solve P(A or B) with a step-by-step guide. Suitable for students in Grades 6-8.

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