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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 - Mathematics

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

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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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 33 Probability
Exercise 33.3 | Q 41.2 | Page 48
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