A bag contains 4 white,5 red,and 6 black ball | vedclass

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(A) Total number of balls $= 4 + 5 + 6 = 15$.Number of ways to draw $2$ balls from $15$ is $^{15}C_2 = \frac{15 	imes 14}{2 	imes 1} = 105$.We want

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$\frac{44}{105}$

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(A) Total number of balls $= 4 + 5 + 6 = 15$.Number of ways to draw $2$ balls from $15$ is $^{15}C_2 = \frac{15 	imes 14}{2 	imes 1} = 105$.We want the probability that one ball is white and the other is non-white (red or black).Number of white balls $= 4$.Number of non-white balls $= 5 + 6 = 11$.Number of ways to choose $1$ white ball and $1$ non-white ball $= ^4C_1 	imes ^{11}C_1 = 4 	imes 11 = 44$.Required probability $= \frac{44}{105}$.

$A$ bag contains $4$ white,$5$ red,and $6$ black balls. If two balls are drawn at random,then the probability that one of them is white is

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