A bag has 6 green, 5 red, 7 blue and 3 pink pencils. One pencil is drawn at random. Find the probability that the pencil is neither green nor pink.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Determine the total number of pencils in the bag.
We have:
- Green pencils = 6
- Red pencils = 5
- Blue pencils = 7
- Pink pencils = 3

Total number of pencils = Green + Red + Blue + Pink
= 6 + 5 + 7 + 3

**Calculation:**
Total = 6 + 5 = 11
Total = 11 + 7 = 18
Total = 18 + 3 = 21

**Total number of pencils = 21**

### Step 2: Identify the favorable outcomes.
We need to find the pencils that are neither green nor pink. Therefore, we will consider only the red and blue pencils.

- Red pencils = 5
- Blue pencils = 7

**Calculation:**
Favorable outcomes = Red + Blue
= 5 + 7

**Favorable outcomes = 12**

### Step 3: Calculate the probability.
Probability is calculated using the formula:
\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} \]

Substituting the values we found:
\[ \text{Probability} = \frac{12}{21} \]

### Step 4: Simplify the probability.
To simplify \( \frac{12}{21} \), we can divide both the numerator and the denominator by their greatest common divisor (GCD), which is 3.

**Calculation:**
\[ \frac{12 \div 3}{21 \div 3} = \frac{4}{7} \]

### Final Answer:
The probability that the pencil drawn is neither green nor pink is \( \frac{4}{7} \).

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