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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.
Solution
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 = \[\frac{n\left( E \right)}{n\left( S \right)} = \frac{^{6}{}{C}_1 \times ^{4}{}{C}_1 \times^{8}{}{C}_1}{^{18}{}{C}_3}\]
\[= \frac{6 \times 4 \times 8}{\frac{18 \times 17 \times 16}{3 \times 2}}\]
\[ = \frac{6 \times 4 \times 8}{6 \times 17 \times 8}\]
\[ = \frac{4}{17}\]
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