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

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(C) The second ball drawn can be white in two mutually exclusive cases:Case $(i)$: The first ball is white and the second ball is white.Probability $=

Vedclass · Accepted Answer

$\frac{4}{7}$

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(C) The second ball drawn can be white in two mutually exclusive cases:Case $(i)$: The first ball is white and the second ball is white.Probability $= \frac{4}{7} 	imes \frac{3}{6} = \frac{12}{42} = \frac{2}{7}$.Case $(ii)$: The first ball is black and the second ball is white.Probability $= \frac{3}{7} 	imes \frac{4}{6} = \frac{12}{42} = \frac{2}{7}$.Therefore,the total probability that the second ball is white is the sum of the probabilities of these two cases:Total Probability $= \frac{2}{7} + \frac{2}{7} = \frac{4}{7}$.

$A$ bag contains $3$ black and $4$ white balls. Two balls are drawn one by one at random without replacement. The probability that the second drawn ball is white,is

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