A bag contains 5 blue and 4 black balls. Three balls are drawn at random. What is the probability that 2 are blue and 1 is black?

Let S be the sample space and E be the event of drawing 3 balls out of which 2 are blue and 1 is back .
Then, n (S)= Number of ways of drawing 3 balls out of 9=⁹c₃
= (9 x 8 x 7) / (3 x 2 x 1)
= 84 and
n(E) = Number of ways of drawing 2 balls out of 5 and 1 ball out of 4 .
= ⁵C₂ ⁴C₁₁
= (5 x 4) / (2 x 1) + 4 = 14
∴ P(E) = n(E)/n(S) = 14/84 = 1/6