A group of five persons is formed from five boys and four girls. Find the probability that there are at least two girls in the group?

Total number of possible ways = 

$9C5 = \frac{9 x 8 x 7 x 6 x 5}{5 x 4 x 3 x 2 x 1} = 126$

Number of favorable cases = 

$4C2 x 5C3 + 4C3 x 5C2 + 4C4 x 5C1\frac{4 x 3}{2 x 1} x \frac{5 x 4 x 3}{3 x 2 x 1} + \frac{4 x 3 x 2}{3 x 2 x 1} x \frac{5 x 4 }{2 x 1} + 1 x 5= 6 x 10 + 4 x 10 + 5= 60 + 40 + 5= 105$

Therefore, required probability = 105/126 = 5/6