There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls ?

Total number of balls in the bag = 8 + 4 + 5 = 17

P(S) = Total possible outcomes

= Selecting 5 balls at random out of 17

=> $$P(S) = C^{17}_5 = \frac{17 \times 16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4 \times 5}$$

= $$6188$$

P(E) = Favorable outcomes

= Selecting 2 brown, 1 orange and 2 black balls.

=> $$P(E) = C^8_2 \times C^4_1 \times C^5_2$$

= $$\frac{8 \times 7}{1 \times 2} \times 4 \times \frac{5 \times 4}{1 \times 2}$$

= $$28 \times 4 \times 10 = 1120$$

$$\therefore$$ Required probability = $$\frac{P(E)}{P(S)}$$ 

= $$\frac{1120}{6188} = \frac{280}{1547}$$