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The probability of drawing two red balls in succession from a bag containing 3 red balls and 4 black balls when the ball that is drawn first is replaced is

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9/49

1/7

1/6

16/81

answer is A.

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Detailed Solution

A= drawing first red ball

B= drawing second red ball

Required probability $P (A \cap B) = P (A) P (B / A) = \frac{3}{7} \times \frac{3}{7} = \frac{9}{49}$

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