Three are 8 brown balls ,4 orange balls and 5 blacks balls in a bag .Five balls are chosen at random. What is the probability of their being 2 brown balls ,I orange ball and 2 black balls ?

Options

A. `(191)/(1547)`
B. `(180)/(1547)`
C. `(280)/(1547)`
D. `(189)/(1547)`

Explanation

To solve the problem of finding the probability of selecting 2 brown balls, 1 orange ball, and 2 black balls from a bag containing 8 brown balls, 4 orange balls, and 5 black balls, we will follow these steps:

### Step-by-Step Solution:

1. **Determine the Total Number of Balls**:
- Brown balls = 8
- Orange balls = 4
- Black balls = 5
- Total balls = 8 + 4 + 5 = 17

2. **Calculate the Total Outcomes**:
- We need to choose 5 balls from the total of 17 balls.
- The total number of ways to choose 5 balls from 17 is given by the combination formula:
\[ \text{Total outcomes} = \binom{17}{5} = \frac{17!}{5!(17-5)!} = \frac{17!}{5! \cdot 12!} \]
- Calculating this:
\[ \binom{17}{5} = \frac{17 \times 16 \times 15 \times 14 \times 13}{5 \times 4 \times 3 \times 2 \times 1} = 6188 \]

3. **Determine the Favorable Outcomes**:
- We need to find the number of ways to choose 2 brown balls, 1 orange ball, and 2 black balls.
- The number of ways to choose 2 brown balls from 8:
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]
- The number of ways to choose 1 orange ball from 4:
\[ \binom{4}{1} = 4 \]
- The number of ways to choose 2 black balls from 5:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

4. **Calculate the Total Favorable Outcomes**:
- Multiply the number of ways to choose each type of ball:
\[ \text{Favorable outcomes} = \binom{8}{2} \times \binom{4}{1} \times \binom{5}{2} = 28 \times 4 \times 10 = 1120 \]

5. **Calculate the Probability**:
- The probability of selecting 2 brown balls, 1 orange ball, and 2 black balls is given by:
\[ P(\text{2 brown, 1 orange, 2 black}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{1120}{6188} \]
- Simplifying this fraction:
\[ P = \frac{1120 \div 4}{6188 \div 4} = \frac{280}{1547} \]

### Final Answer:
The probability of selecting 2 brown balls, 1 orange ball, and 2 black balls is \(\frac{280}{1547}\).