A bag contains 3 red and 3 green balls and a person draws out 3 at random. He then drops 3 blue balls into the bag and again draws out 3 at random. The chance that the 3 later balls being all of different colors is `15 %` b. `20 %` c. `27 %` d. `40 %`

Options

A. `15%`
B. `20%`
C. `27%`
D. `40%`

Explanation

To solve the problem step by step, we need to calculate the probability that after drawing 3 balls from a bag containing red, green, and blue balls, all drawn balls are of different colors.

### Step 1: Understand the initial setup
Initially, the bag contains:
- 3 Red balls
- 3 Green balls

Total = 6 balls

### Step 2: Draw 3 balls from the bag
We need to consider the different combinations of drawing 3 balls from the initial 6. The total ways to choose 3 balls from 6 is calculated using the combination formula:

\[ \text{Total ways to choose 3 balls} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = 20 \]

### Step 3: Add 3 Blue balls to the bag
After drawing 3 balls, the person adds 3 Blue balls to the bag. The composition of the bag now depends on the colors of the balls drawn.

### Step 4: Calculate the cases for drawing balls of different colors
We want to find the probability that the next 3 balls drawn are one of each color (1 Red, 1 Green, and 1 Blue).

#### Case Analysis
1. **Case 1**: 1 Red, 1 Green, and 1 Blue
- The number of ways to choose 1 Red from the remaining 2 Red, 1 Green from the remaining 2 Green, and 1 Blue from the 3 Blue:
\[ \text{Ways} = \binom{2}{1} \times \binom{2}{1} \times \binom{3}{1} = 2 \times 2 \times 3 = 12 \]

2. **Case 2**: 2 Red and 1 Green
- The number of ways to choose 2 Red from the remaining 2 Red, and 1 Green from the remaining 2 Green:
\[ \text{Ways} = \binom{2}{2} \times \binom{2}{1} \times \binom{3}{0} = 1 \times 2 \times 1 = 2 \]

3. **Case 3**: 1 Red and 2 Green
- The number of ways to choose 1 Red from the remaining 2 Red, and 2 Green from the remaining 2 Green:
\[ \text{Ways} = \binom{2}{1} \times \binom{2}{2} \times \binom{3}{0} = 2 \times 1 \times 1 = 2 \]

### Step 5: Total ways to draw 3 balls after adding Blue
The total ways to choose 3 balls from the bag after adding 3 Blue balls (total 9 balls now):
\[ \text{Total ways to choose 3 balls} = \binom{9}{3} = 84 \]

### Step 6: Calculate the probability
The total favorable outcomes for drawing 1 Red, 1 Green, and 1 Blue are from Case 1, which is 12.

Thus, the probability \( P \) that the 3 balls drawn are all of different colors is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{12}{84} = \frac{2}{14} = \frac{1}{7} \]

### Step 7: Convert to percentage
To convert the probability into a percentage:
\[ P = \frac{1}{7} \times 100 \approx 14.29\% \]

### Conclusion
The probability that the 3 balls drawn are all of different colors is approximately 14.29%. Since this does not match any of the provided options, we can conclude that the calculations may need to be verified or the options may have been misrepresented.