Question

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?
\\(\frac{12}{190}\\)
\\(\frac{33}{95}\\)
\\(\frac{62}{95}\\)
\\(\frac{178}{190}\\)

Explanation:

Step1: Calculate total number of ways to choose 2 students

The total number of students is $8 + 12=20$. The number of ways to choose 2 students out of 20 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 20$ and $r = 2$. So $C(20,2)=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times1}=190$.

Step2: Calculate number of ways to choose 2 girls

The number of ways to choose 2 girls out of 12 is $C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66$.

Step3: Calculate probability of choosing 2 girls

The probability of choosing 2 girls, $P(\text{2 girls})=\frac{C(12,2)}{C(20,2)}=\frac{66}{190}=\frac{33}{95}$.

Step4: Calculate probability of not choosing 2 girls

The probability of not choosing 2 girls is $P = 1 - P(\text{2 girls})=1-\frac{33}{95}=\frac{95 - 33}{95}=\frac{62}{95}$.

Answer:

$\frac{62}{95}$