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: Calcular el total de estudiantes

Hay 8 hombres y 12 mujeres, entonces el total de estudiantes es $8 + 12=20$.

Step2: Calcular el número de maneras de elegir 2 estudiantes

Usando la fórmula de combinaciones $C(n,r)=\frac{n!}{r!(n - r)!}$, donde $n = 20$ y $r=2$. Entonces $C(20,2)=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times1}=190$.

Step3: Calcular el número de maneras de elegir 2 mujeres

Aquí $n = 12$ y $r = 2$. Entonces $C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66$.

Step4: Calcular la probabilidad de elegir 2 mujeres

$P(\text{2 mujeres})=\frac{C(12,2)}{C(20,2)}=\frac{66}{190}=\frac{33}{95}$.

Step5: Calcular la probabilidad de no elegir 2 mujeres

$P(\text{no 2 mujeres})=1 - P(\text{2 mujeres})=1-\frac{33}{95}=\frac{95 - 33}{95}=\frac{62}{95}$.

Answer:

C. $\frac{62}{95}$