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(B) Total number of balls $= 4 + 5 + 6 = 15$.We need to draw $3$ balls of different colours,which means one red,one blue,and one yellow ball.The numbe

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$\frac{24}{91}$

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(B) Total number of balls $= 4 + 5 + 6 = 15$.We need to draw $3$ balls of different colours,which means one red,one blue,and one yellow ball.The number of ways to choose $1$ red,$1$ blue,and $1$ yellow ball is $\binom{4}{1} 	imes \binom{5}{1} 	imes \binom{6}{1} = 4 	imes 5 	imes 6 = 120$.The total number of ways to draw $3$ balls from $15$ is $\binom{15}{3} = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 5 	imes 7 	imes 13 = 455$.The required probability is $\frac{120}{455} = \frac{24}{91}$.

$3$ balls are drawn one after the other without replacement from an urn containing $4$ red,$5$ blue and $6$ yellow balls. The probability of getting three different coloured balls is

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