Question

Finish Color Matte Finish Glossy Finish TOTAL Red 8 9 17 Blue 2 4 6 Yellow 0 7 7 TOTAL 10 20 30 Find the P(blue or matte) . Provide the answer in fraction form.

Answer 1

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

\[
\boxed{\frac{7}{15}}
\]