two students from a group of eight boys and 12 girls are sent to represent the school in a parade. if the…

two students from a group of eight boys and 12 girls are sent to represent the school in a parade. if the students are chosen at random, what is the probability that the students chosen are not both girls?\n\\(\\frac{12}{190}\\)\n\\(\\frac{33}{95}\\)\n\\(\\frac{62}{95}\\)\n\\(\\frac{178}{190}\\)
Answer
Explanation:
Step1: Calculate total number of students
There are 8 boys and 12 girls, so total students $n = 8 + 12=20$.
Step2: Calculate total number of ways to choose 2 students
Using combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, here $n = 20$ and $r = 2$. So $C(20,2)=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times1}=190$.
Step3: Calculate number of ways to choose 2 girls
Number of girls is 12, using combination formula with $n = 12$ and $r = 2$. So $C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66$.
Step4: Calculate probability of choosing 2 girls
$P(\text{both girls})=\frac{C(12,2)}{C(20,2)}=\frac{66}{190}=\frac{33}{95}$.
Step5: Calculate probability of not choosing both girls
$P(\text{not both girls}) = 1 - P(\text{both girls})=1-\frac{33}{95}=\frac{95 - 33}{95}=\frac{62}{95}$.
Answer:
$\frac{62}{95}$