In the class, $ Boys = Girls + 2 $
& $ Boys + Girls = 12 $
$\color{blue}{\text{∴ Boys = 7 & Girls = 5}}$
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 = ${^5C_2} \times {^7C_1}$
Choosing $3$ girls from $5$ girls = ${^5C_3}$
$\color{maroon}{\text{∴ Probability that the group accompanying the teacher contains more girls than boys }}$
$=\color{purple}{\dfrac{\text{number of favorable ways}}{\text{total number of ways}}}$
$=\dfrac{{^5C_2 } \times {^7C_1} + {^5C_3 }}{{^{12}C_3}}$
$=\dfrac{(10)\times(7)+(10)}{220}$
$=\dfrac{80}{220}=\dfrac{4}{11}$
$\color{green}{\text{Hence, the answer is}}\color{purple} \ {\text{None of these}}$
This was marks to all in GATE