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Question
An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that one ball is red and the other is black
Solution
Out of 15 balls, two balls can be drawn in 15C2 ways.
∴ Total number of elementary events = 15C2 = 105
Out of three red balls, one red ball can be drawn in 3C1 ways; and out of five black balls, one black ball can be drawn in 5C1 ways.
Therefore, one red and one black can be drawn in 3C1× 5C1 ways.
∴ Favourable number of ways = 3C1× 5C1 = 3× 5 = 15
Hence, required probability = \[\frac{15}{105} = \frac{1}{7}\]
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