A bag contains 6 white and 4 black balls. Two | vedclass

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(D) Total number of balls $= 6 + 4 = 10$. Number of ways to select $2$ balls out of $10 = {}^{10}C_2 = \frac{10 	imes 9}{2 	imes 1} = 45$. Number of

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$7/15$

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(D) Total number of balls $= 6 + 4 = 10$. Number of ways to select $2$ balls out of $10 = {}^{10}C_2 = \frac{10 	imes 9}{2 	imes 1} = 45$. Number of ways to select $2$ white balls from $6 = {}^{6}C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$. Number of ways to select $2$ black balls from $4 = {}^{4}C_2 = \frac{4 	imes 3}{2 	imes 1} = 6$. Total favorable outcomes $= 15 + 6 = 21$. $	herefore$ The required probability $= \frac{21}{45} = \frac{7}{15}$.

$A$ bag contains $6$ white and $4$ black balls. Two balls are drawn at random. The probability that they are of the same colour is:

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