There are 12 boys and 8 girls in a tuition centre. If three of them scored first mark, then what is the probability that one of the three is a girl and the other two are boys?

Solution(By Examveda Team)

Total number of students = 20
Let S be the sample space
Then, n(S) = number of ways of three scored first mark
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{20}{C_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20 \times 19 \times 18}}{{2 \times 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 19 \times 3 \cr} $$
Let, E be the event of 1 girl and 2 boys
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^8{C_1} \times {}^{12}{C_2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 12 \times 11}}{{1 \times 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \times 6 \times 11 \cr} $$
Now, the required probability
$$\eqalign{ & = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & = \frac{{8 \times 6 \times 11}}{{20 \times 19 \times 3}} \cr & = \frac{{44}}{{95}} \cr} $$