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Question
An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting one red and one blue ball.
Solution
\[\text{ Total balls = 4 red balls + 7 blue balls = 11 balls } \] \[ P\left( \text{ one red and one blue } \right) = P\left( \text{ first red and second blue } \right) + P\left( \text{ first blue and second red } \right)\]
\[ = \frac{4}{11} \times \frac{7}{11} + \frac{7}{11} \times \frac{4}{11}\]
\[ = \frac{28}{121} + \frac{28}{121}\]
\[ = \frac{56}{121}\]
\[\text{ Disclaimer: In the question,instead of black ballsit should be blue balls.} \]
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