In a class there are 10 boys and 3 girls . A game was organized , find the probability coming at least one girl in first three positions when all students participated in that game .

Options

A. `(3)/(10)`
B. `(83)/(143)`
C. `(157)/(286)`
D. `(167)/(289)`

Answer

To solve the problem of finding the probability of having at least one girl in the first three positions when 10 boys and 3 girls participate in a game, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Students**:
- There are 10 boys and 3 girls in the class.
- Total number of students = 10 boys + 3 girls = 13 students.

2. **Calculate the Total Ways to Arrange Students**:
- We need to find the total number of ways to choose 3 students from 13.
- This can be calculated using the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
- Therefore, the total ways to choose 3 students from 13 is:
\[ C(13, 3) = \frac{13!}{3!(13-3)!} = \frac{13!}{3! \cdot 10!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286. \]

3. **Calculate the Ways to Arrange Only Boys**:
- Now, we calculate the number of ways to choose 3 boys from the 10 boys.
- Using the combination formula again:
\[ C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120. \]

4. **Calculate the Probability of Choosing Only Boys**:
- The probability of choosing only boys in the first three positions is given by the ratio of the number of ways to choose boys to the total ways to choose students:
\[ P(\text{only boys}) = \frac{C(10, 3)}{C(13, 3)} = \frac{120}{286}. \]

5. **Calculate the Probability of At Least One Girl**:
- To find the probability of having at least one girl in the first three positions, we can use the complement rule:
\[ P(\text{at least one girl}) = 1 - P(\text{only boys}). \]
- Substituting the values we calculated:
\[ P(\text{at least one girl}) = 1 - \frac{120}{286} = \frac{286 - 120}{286} = \frac{166}{286}. \]

6. **Simplify the Probability**:
- We can simplify \( \frac{166}{286} \) by finding the greatest common divisor (GCD):
- The GCD of 166 and 286 is 2.
- Therefore, we simplify:
\[ \frac{166 \div 2}{286 \div 2} = \frac{83}{143}. \]

### Final Answer:
The probability of having at least one girl in the first three positions is \( \frac{83}{143} \).