A class of twelve children has two more boys than girls. A group of three children are randomly picked from this class to accompany the teacher on a field trip. What is the probability that the group accompanying the teacher contains more girls than boys?

1. $0 \\$
2. $\frac{325}{864} \\$
3. $\frac{525}{864} \\$
4. $\frac{5}{12}$

edited ago

In the class, Boys = Girls + 2

& Boys + Girls = 12

∴ Boys = 7 & Girls = 5

Now, We have to choose 3 students from 12 students which can be done in  ${^{12}C_3}$

But among these 3 students, girls should be more than the boys

It can be done in 2 ways

either in the field trip, there are 2 girls and 1 boys OR there are 3 girls

∴  Choosing 2 girls from 5 girls AND choosing 1 boys from 7 boys = ${^{5}C_2}$ * ${^{7}C_1}$

Choosing 3 girls from 5 girls =  ${^{5}C_3}$

Probability that the group accompanying the teacher contains more girls than boys = $\dfrac{number of favorable ways}{total number of ways}$

= $\dfrac{{5 \choose 2} * {7 \choose 1} + {5 \choose 3}}{{12 \choose 3}}$

= $\dfrac{(10)*(7)+(10)}{220}$

= $\dfrac{80}{220}$

= 0.36 ≅ $\dfrac{325}{864}$

Hence, the answer is option B)

by (720 points)