Out of $10$ meritorious students in a school, there are $6$ girls and $4$ boys. A team of $4$ students is selected at random for a quiz program. Find the probability that there are at least $2$ girls.
$4$ students can be selected from $10$ students in
${^{10}}{C}_{4} = \dfrac{10!}{6! \times 4!} = \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$ ways
$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 210$
Let $A$ be the event of selecting $4$ students so that there are at least $2$ girls i.e. selecting $2$ girl students OR $3$ girl students OR $4$ girl students
Now,
$2$ girls can be selected from $6$ girls in ${^{6}}{C}_{2} = \dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2 \times 1} = 15$ ways
$\hspace{4cm}$ AND
$2$ boys can be selected from $4$ boys in ${^{4}}{C}_{2} = \dfrac{4!}{2! \times 2!} = \dfrac{4 \times 3}{2 \times 1} = 6$ ways
$\therefore \;$ $2$ girls and $2$ boys can be selected in $15 \times 6 = 90$ ways
$\hspace{4cm}$ OR
$3$ girls can be selected from $6$ girls in ${^{6}}{C}_{3} = \dfrac{6!}{3! \times 3!} = \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways
$\hspace{4cm}$ AND
$1$ boy can be selected from $4$ boys in ${^{4}}{C}_{1} = 4$ ways
$\therefore \;$ $2$ girls and $1$ boy can be selected in $20 \times 4 = 80$ ways
$\hspace{4cm}$ OR
$4$ girls can be selected from $6$ girls in ${^{6}}{C}_{4} = \dfrac{6!}{2! \times 4!} = \dfrac{6 \times 5}{2 \times 1} = 15$ ways
$\therefore \;$ Number of elements in $A = n\left(A\right) = 90 + 80 + 15 = 185$ ways
$\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n\left(S\right)} = \dfrac{185}{210} = \dfrac{37}{42}$