A basket contains 3 blue and 4 red balls. If | vedclass

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(B) Total number of balls $= 3 + 4 = 7$.Total ways to draw $3$ balls from $7$ is given by ${}^{7}C_{3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes

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$\frac{1}{7}$

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(B) Total number of balls $= 3 + 4 = 7$.Total ways to draw $3$ balls from $7$ is given by ${}^{7}C_{3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.Case $1$: All $3$ balls are blue. The number of ways to choose $3$ blue balls from $3$ is ${}^{3}C_{3} = 1$.Case $2$: All $3$ balls are red. The number of ways to choose $3$ red balls from $4$ is ${}^{4}C_{3} = 4$.Since these are mutually exclusive events,the total number of favorable outcomes is $1 + 4 = 5$.Required probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{5}{35} = \frac{1}{7}$.

$A$ basket contains $3$ blue and $4$ red balls. If $3$ balls are drawn at random from the basket,then what is the probability that all the $3$ are either blue or red?

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