A bag contains 4 red gumballs, 7 green gumballs, 2 white gumballs , and 5 blue gumballs. How many additional white gumballs nust be added to the 18 gumballls is the bag so thaat the probability of drawing a gumball that is not white is `2/3` ?

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to determine how many additional white gumballs (let's denote this number as \( x \)) must be added to the existing gumballs in the bag so that the probability of drawing a gumball that is not white is \( \frac{2}{3} \).

### Step-by-Step Solution:

1. **Identify the current number of gumballs:**
- Red gumballs: 4
- Green gumballs: 7
- White gumballs: 2
- Blue gumballs: 5

The total number of gumballs currently in the bag is:
\[ 4 + 7 + 2 + 5 = 18 \]

2. **Set up the equation for total gumballs after adding \( x \) white gumballs:**
After adding \( x \) white gumballs, the total number of gumballs becomes:
\[ 18 + x \]

3. **Calculate the number of favorable outcomes (gumballs that are not white):**
The number of gumballs that are not white is:
\[ 4 \text{ (red)} + 7 \text{ (green)} + 5 \text{ (blue)} = 16 \]

4. **Set up the probability equation:**
The probability of drawing a gumball that is not white is given by:
\[ \text{Probability} = \frac{\text{Number of non-white gumballs}}{\text{Total number of gumballs}} = \frac{16}{18 + x} \]
We want this probability to equal \( \frac{2}{3} \):
\[ \frac{16}{18 + x} = \frac{2}{3} \]

5. **Cross-multiply to solve for \( x \):**
Cross-multiplying gives:
\[ 16 \cdot 3 = 2 \cdot (18 + x) \]
Simplifying this:
\[ 48 = 36 + 2x \]

6. **Isolate \( x \):**
Subtract 36 from both sides:
\[ 48 - 36 = 2x \]
\[ 12 = 2x \]
Divide both sides by 2:
\[ x = 6 \]

### Conclusion:
The number of additional white gumballs that must be added to the bag is \( \boxed{6} \).