Question

A bag contains 6 red, 4 white and 8 blue balls. if three balls are drawn at random, find the probability that one is red, one is white and one is blue.

Answer 1

Total number of balls = 6 + 4 + 8 = 18
Total number of elementary events, n(S) = ¹⁸C₃
Let E be the event of favourable outcomes.
Here, E = getting one red, one white and one blue ball
So, favourable number of elementary events, n(E) = ⁶C₁ ×⁴C₁ × ⁸C₁
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^{6}C_1\times {}^{4}C_1\times {}^{8}C_1}{{}^{18}C_3}$
$$\begin{gathered} =\frac{6\times 4\times 8}{{\displaystyle \frac{18\times 17\times 16}{3\times 2}}} \\ =\frac{6\times 4\times 8}{6\times 17\times 8} \\ =\frac{4}{17} \end{gathered}$$