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

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(C) Total number of balls $= 6 + 4 = 10$. Number of ways to draw $2$ balls from $10$ is given by $^{10}C_2 = \frac{10 	imes 9}{2 	imes 1} = 45$. For

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$\frac{7}{15}$

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(C) Total number of balls $= 6 + 4 = 10$. Number of ways to draw $2$ balls from $10$ is given by $^{10}C_2 = \frac{10 	imes 9}{2 	imes 1} = 45$. For the balls to be of the same colour,they must either be both white or both black. Number of ways to draw $2$ white balls from $6$ is $^{6}C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$. Number of ways to draw $2$ black balls from $4$ is $^{4}C_2 = \frac{4 	imes 3}{2 	imes 1} = 6$. Total favorable outcomes $= 15 + 6 = 21$. 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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