Sure, let's solve this step-by-step.
First, we need to find the total number of students in the group. We are given that there are 8 boys and 12 girls.
[tex]\[ \text{Total number of students} = 8 + 12 = 20 \][/tex]
Next, we need to calculate the total number of ways to choose 2 students out of these 20. This is a combination problem where we choose 2 students from 20, which can be represented as [tex]\(\binom{20}{2}\)[/tex].
[tex]\[ \binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20 \times 19}{2 \times 1} = 190 \][/tex]
Now, we need to find the number of ways to choose 2 girls out of the 12 girls. This again is a combination problem represented as [tex]\(\binom{12}{2}\)[/tex].
[tex]\[ \binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66 \][/tex]
So, there are 66 ways to choose 2 girls out of the 12 girls.
To find the number of ways to choose students such that they are not both girls, we subtract the number of ways to choose 2 girls from the total number of ways to choose 2 students:
[tex]\[ \text{Number of ways to choose students not both girls} = \binom{20}{2} - \binom{12}{2} = 190 - 66 = 124 \][/tex]
The probability that the students chosen are not both girls is the ratio of the number of favorable outcomes to the total number of outcomes:
[tex]\[ \text{Probability} = \frac{\text{Number of ways to choose students not both girls}}{\text{Total number of ways to choose 2 students}} = \frac{124}{190} = \frac{62}{95} \][/tex]
Therefore, the probability that the students chosen are not both girls is:
[tex]\[ \boxed{\frac{62}{95}} \][/tex]
