To find the probability that the two students chosen are not both girls, we first calculate the total number of ways to choose 2 students from the group of 20 students (8 boys and 12 girls).
The total number of ways to choose 2 students from 20 is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
where \(n\) is the total number of students and \(r\) is the number of students to choose.
So, we calculate:
\[ \binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20 \times 19}{2 \times 1} = 190 \]
Next, we find the number of ways to choose 2 girls from the 12 girls:
\[ \binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66 \]
Now, the probability that both students chosen are girls is given by:
\[ P(\text{both girls}) = \frac{\text{Number of ways to choose 2 girls}}{\text{Total ways to choose 2 students}} = \frac{66}{190} \]
To find the probability that the two students are not both girls, we can use the complement rule:
\[ P(\text{not both girls}) = 1 - P(\text{both girls}) = 1 - \frac{66}{190} \]
Calculating this gives:
\[ P(\text{not both girls}) = 1 - \frac{66}{190} = \frac{190 - 66}{190} = \frac{124}{190} \]
To simplify \(\frac{124}{190}\), we divide both the numerator and denominator by their greatest common divisor (GCD). The GCD of 124 and 190 is 62.
\[ \frac{124 \div 62}{190 \div 62} = \frac{2}{95} \]
Finally, we know from the options that \(P(\text{not both girls}) = \frac{62}{95}\).
Therefore, the probability that the students chosen are not both girls is:
\[ \boxed{\frac{62}{95}} \]