A bag contains 3 red,4 white and 5 black ball | vedclass

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(A) Total number of balls $= 3 + 4 + 5 = 12$.Number of ways to draw $3$ balls out of $12$ is given by $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	im

Vedclass · Accepted Answer

$\frac{3}{11}$

Vedclass · Answer

(A) Total number of balls $= 3 + 4 + 5 = 12$.Number of ways to draw $3$ balls out of $12$ is given by $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.For the balls to be of different colours,we must draw $1$ red,$1$ white,and $1$ black ball.Number of favorable outcomes $= ^3C_1 	imes ^4C_1 	imes ^5C_1 = 3 	imes 4 	imes 5 = 60$.Probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{60}{220} = \frac{6}{22} = \frac{3}{11}$.

$A$ bag contains $3$ red,$4$ white and $5$ black balls. Three balls are drawn at random. The probability of the balls being of different colours is

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