Anurn contains 6 red,4 blue ,2 green and 3yellow marbles
If four marbles are picked at random , what is the probability that one is green , two are blue and one is red ?

Options

A. `(24)/(455)`
B. `(13)/(35)`
C. `(11)/(15)`
D. `(1)/(3)`

Explanation

To solve the problem of finding the probability that one marble is green, two are blue, and one is red when picking four marbles from Anurban's collection, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the total number of marbles**:
- Red marbles = 6
- Blue marbles = 4
- Green marbles = 2
- Yellow marbles = 3
- **Total marbles = 6 + 4 + 2 + 3 = 15**

2. **Calculate the total number of ways to choose 4 marbles from 15**:
- The total number of ways to choose 4 marbles from 15 is given by the combination formula:
\[ \text{Total ways} = \binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \]

3. **Calculate the number of favorable outcomes**:
- We want 1 green, 2 blue, and 1 red marble.
- The number of ways to choose 1 green marble from 2:
\[ \binom{2}{1} = 2 \]
- The number of ways to choose 2 blue marbles from 4:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
- The number of ways to choose 1 red marble from 6:
\[ \binom{6}{1} = 6 \]
- Therefore, the total number of favorable outcomes is:
\[ \text{Favorable outcomes} = \binom{2}{1} \times \binom{4}{2} \times \binom{6}{1} = 2 \times 6 \times 6 = 72 \]

4. **Calculate the probability**:
- The probability of picking 1 green, 2 blue, and 1 red marble is given by:
\[ P(\text{1 green, 2 blue, 1 red}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{72}{1365} \]

5. **Simplify the probability**:
- The fraction can be simplified if possible. In this case, it is already in its simplest form.

### Final Answer:
The probability that one marble is green, two are blue, and one is red is:
\[ \frac{72}{1365} \]