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)