To find \( P(\text{blue or matte}) \), we can use the principle of inclusion-exclusion:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
where:
- \( A \) is the event of selecting a blue item.
- \( B \) is the event of selecting a matte finish item.
Step 1: Calculate \( P(A) \) (Probability of choosing blue)
The total number of items is \( 30 \).
The number of blue items is \( 6 \).
\[ P(A) = \frac{\text{Number of blue items}}{\text{Total items}} = \frac{6}{30} = \frac{1}{5} \]
Step 2: Calculate \( P(B) \) (Probability of choosing matte finish)
The number of matte finish items is \( 10 \).
\[ P(B) = \frac{\text{Number of matte finish items}}{\text{Total items}} = \frac{10}{30} = \frac{1}{3} \]
Step 3: Calculate \( P(A \cap B) \) (Probability of choosing blue and matte)
The number of blue matte items is \( 2 \) (from the table provided).
\[ P(A \cap B) = \frac{\text{Number of blue matte items}}{\text{Total items}} = \frac{2}{30} = \frac{1}{15} \]
Step 4: Calculate \( P(A \cup B) \) (Probability of choosing blue or matte)
Now we can use the inclusion-exclusion principle:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the values we found:
\[ P(A \cup B) = \frac{1}{5} + \frac{1}{3} - \frac{1}{15} \]
Step 5: Find a common denominator
The least common multiple of \( 5, 3, \) and \( 15 \) is \( 15 \).
Now convert each fraction to have a denominator of \( 15 \):
- \( P(A) = \frac{1}{5} = \frac{3}{15} \)
- \( P(B) = \frac{1}{3} = \frac{5}{15} \)
- \( P(A \cap B) = \frac{1}{15} \)
Step 6: Combine these fractions
\[ P(A \cup B) = \frac{3}{15} + \frac{5}{15} - \frac{1}{15} = \frac{3 + 5 - 1}{15} = \frac{7}{15} \]
Final Answer
The probability \( P(\text{blue or matte}) \) is
\[ \boxed{\frac{7}{15}} \]