A bag contains 5 red and 4 green balls. If fo | vedclass

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(B) Total number of ways to draw $4$ balls from $9$ balls is $n(S) = ^9C_4 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.

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$10/21$

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(B) Total number of ways to draw $4$ balls from $9$ balls is $n(S) = ^9C_4 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.Number of ways to choose $2$ red balls from $5$ red balls is $n(A_1) = ^5C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$.Number of ways to choose $2$ green balls from $4$ green balls is $n(A_2) = ^4C_2 = \frac{4 	imes 3}{2 	imes 1} = 6$.Total favorable outcomes $n(A) = n(A_1) 	imes n(A_2) = 10 	imes 6 = 60$.The probability $P(A) = \frac{n(A)}{n(S)} = \frac{60}{126} = \frac{10}{21}$.

$A$ bag contains $5$ red and $4$ green balls. If four balls are drawn at random,what is the probability that two balls are red and two balls are green?

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