Sure, let's solve this step-by-step!
We are given a group consisting of 8 boys and 12 girls, making a total of:
[tex]\[ \text{Total students} = 8 + 12 = 20 \][/tex]
We need to find the probability that when 2 students are chosen at random, they are not both girls.
First, let's calculate the total number of ways to choose 2 students out of the 20 students. We use the combination formula:
[tex]\[ \binom{20}{2} = \frac{20!}{2!(20-2)!} = 190 \][/tex]
Next, let's determine the number of ways to choose 2 girls out of the 12 girls:
[tex]\[ \binom{12}{2} = \frac{12!}{2!(12-2)!} = 66 \][/tex]
Now, to find the number of ways to choose at least one boy, we subtract the number of ways to choose 2 girls from the total number of ways to choose 2 students:
[tex]\[ \text{Ways to choose at least one boy} = 190 - 66 = 124 \][/tex]
Therefore, the probability that the students chosen are not both girls is the number of ways to choose at least one boy divided by the total number of ways to choose any two students:
[tex]\[ P(\text{at least one boy}) = \frac{124}{190} = \frac{62}{95} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\frac{62}{95}} \][/tex]
