A bucket contains 8red,3 blue and 5 green marbles .
If 4 marbles are drawn at random , what is the probabaility that 2 are and 2 are blue ?

Options

A. `(11)/(16)`
B. `(3)/(16)`
C. `(11)/(72)`
D. `(3)/(65)`

Explanation

To solve the problem of finding the probability that when drawing 4 marbles from a bucket containing 8 red, 3 blue, and 5 green marbles, exactly 2 are red and 2 are blue, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Total Number of Marbles:**
- Total red marbles = 8
- Total blue marbles = 3
- Total green marbles = 5
- Total marbles = 8 + 3 + 5 = 16

**Hint:** Always start by calculating the total number of items in the set.

2. **Calculate the Total Number of Ways to Draw 4 Marbles:**
- The total number of ways to choose 4 marbles from 16 is given by the combination formula:
\[ \text{Total Outcomes} = \binom{16}{4} = \frac{16!}{4!(16-4)!} = \frac{16!}{4! \cdot 12!} \]
- Calculating this:
\[ \binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820 \]

**Hint:** Use the combination formula to find the number of ways to choose items from a larger set.

3. **Determine the Number of Favorable Outcomes:**
- We want to find the number of ways to choose 2 red marbles from 8 and 2 blue marbles from 3.
- The number of ways to choose 2 red marbles:
\[ \text{Ways to choose 2 red} = \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]
- The number of ways to choose 2 blue marbles:
\[ \text{Ways to choose 2 blue} = \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3}{1} = 3 \]
- Therefore, the total number of favorable outcomes is:
\[ \text{Favorable Outcomes} = \binom{8}{2} \times \binom{3}{2} = 28 \times 3 = 84 \]

**Hint:** Break down the problem into smaller parts by calculating the combinations for each color separately.

4. **Calculate the Probability:**
- The probability of drawing 2 red and 2 blue marbles is given by the ratio of favorable outcomes to total outcomes:
\[ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{84}{1820} \]
- Simplifying this fraction:
\[ \frac{84}{1820} = \frac{21}{455} \]

**Hint:** Always simplify your final answer to its lowest terms for clarity.

5. **Final Answer:**
- The probability that 2 marbles drawn are red and 2 are blue is:
\[ \frac{21}{455} \]