Question

An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is
(a) 5/84
(b) 3/9
(c) 3/7
(d) 7/17

Answer 1

Total number of Balls = 9
Total number of ways of selecting 3 balls ⁹C₃
Since three red colored balls cannot be selected
⇒ number of ways of selecting same colored balls
$$\begin{gathered} ={}^3\mathrm{C}_3+{}^4\mathrm{C}_3 \\ =1+4 \\ =5 \end{gathered}$$
∴ Probability of selecting same color balls

$$\begin{gathered} =\frac{\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{selecting}\mathrm{some}\mathrm{colored}\mathrm{balls}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{selecting}3\mathrm{balls}} \\ =\frac{5}{{}^9\mathrm{C}_3}=\frac{5\times 3!\times 6!}{9!} \\ =\frac{5\times 2\times 3\times 6!}{9\times 8\times 7\times 6!} \\ =\frac{5}{84} \end{gathered}$$
Hence, the correct answer is option A.