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: Find total number of students

Total students = 8 boys + 12 girls = 20 students.

Step2: Calculate total number of ways to choose 2 students

Using combination formula \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 20 \), \( k = 2 \).
\( C(20, 2)=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times1}=190 \).

Step3: Calculate number of ways to choose 2 girls

\( n = 12 \), \( k = 2 \).
\( C(12, 2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66 \).

Step4: Calculate probability of both girls

Probability(both girls) = \( \frac{C(12, 2)}{C(20, 2)}=\frac{66}{190}=\frac{33}{95} \).

Step5: Calculate probability of not both girls

Probability(not both girls) = 1 - Probability(both girls) = \( 1-\frac{33}{95}=\frac{95 - 33}{95}=\frac{62}{95} \).

Answer:

\(\frac{62}{95}\) (corresponding to the option with \(\frac{62}{95}\))