A bag contains 8 red and 4 green balls . Find the probabilty that
three balls are drawn and none of them is red .

Options

A. `68/99`
B. `7/99`
C. `4/495`
D. none of these

Answer

To solve the problem, we need to find the probability that when drawing 3 balls from a bag containing 8 red balls and 4 green balls, none of the drawn balls is red.

### Step-by-Step Solution:

1. **Identify the Total Number of Balls**:
- The bag contains 8 red balls and 4 green balls.
- Total number of balls = 8 (red) + 4 (green) = 12 balls.

2. **Determine the Favorable Outcomes**:
- We want to find the probability that all 3 drawn balls are green.
- The number of green balls = 4.
- We need to choose 3 green balls from these 4. This can be calculated using combinations.

3. **Calculate the Number of Ways to Choose 3 Green Balls**:
- The number of ways to choose 3 balls from 4 is given by the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).
- Here, \( n = 4 \) and \( r = 3 \):
\[ \text{Favorable outcomes} = \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3! \cdot 1!} = \frac{4 \cdot 3!}{3! \cdot 1} = 4 \]

4. **Calculate the Total Outcomes**:
- The total number of ways to choose 3 balls from 12 is:
\[ \text{Total outcomes} = \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3! \cdot 9!} = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} = \frac{1320}{6} = 220 \]

5. **Calculate the Probability**:
- The probability \( P \) that none of the drawn balls is red (i.e., all are green) is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{none red}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{4}{220} = \frac{1}{55} \]

### Final Answer:
The probability that none of the drawn balls is red is \( \frac{1}{55} \).