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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 and two are white
Solution
Out of 18 balls, three balls can be drawn in 18C3 ways.
∴ Total number of elementary events = 18C3 = 816
Out of six red balls, one red ball can be drawn in 6C1 ways.
Out of four white balls, two white balls can be drawn in 4C2 ways .
Therefore, one red and two white balls can be drawn in 6C1 × 4C2 = 6 × 6 = 36 ways
∴ Favourable number of ways = 36
Hence, required probability = \[\frac{36}{816} = \frac{3}{68}\]
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