A box contains 6 black and 4 red balls. Three | vedclass

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(A) Total number of balls = $6 + 4 = 10$.Number of ways to draw $3$ balls from $10$ balls is given by ${}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	im

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$1/2$

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(A) Total number of balls = $6 + 4 = 10$.Number of ways to draw $3$ balls from $10$ balls is given by ${}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.Number of ways to select $1$ red ball from $4$ red balls is ${}^{4}C_1 = 4$.Number of ways to select $2$ black balls from $6$ black balls is ${}^{6}C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$.Number of favourable outcomes = ${}^{4}C_1 	imes {}^{6}C_2 = 4 	imes 15 = 60$.Probability $P(E) = \frac{	ext{Favourable outcomes}}{	ext{Total outcomes}} = \frac{60}{120} = \frac{1}{2}$.

$A$ box contains $6$ black and $4$ red balls. Three balls are drawn at random. What is the probability that one ball is red and the other two are black?

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