A bag contains 4 white and 3 red balls. Two d | vedclass

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(A) Total number of balls = $4$ (white) + $3$ (red) = $7$ balls.Probability of drawing the first red ball = $\frac{3}{7}$.Since the ball is not replac

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(A) Total number of balls = $4$ (white) + $3$ (red) = $7$ balls.Probability of drawing the first red ball = $\frac{3}{7}$.Since the ball is not replaced,the remaining number of balls is $6$,and the remaining number of red balls is $2$.Probability of drawing the second red ball = $\frac{2}{6} = \frac{1}{3}$.Therefore,the probability that both balls are red = $\frac{3}{7} 	imes \frac{1}{3} = \frac{1}{7}$.

$A$ bag contains $4$ white and $3$ red balls. Two draws of one ball each are made without replacement. Then the probability that both the balls are red is (in $/7$)

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