A bag contains 2 red,3 green and 2 blue balls | vedclass

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(A) Total number of balls $= 2 + 3 + 2 = 7$.The number of ways to draw $2$ balls out of $7$ is given by ${}^{7}C_{2} = \frac{7 	imes 6}{1 	imes 2} =

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$\frac{10}{21}$

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(A) Total number of balls $= 2 + 3 + 2 = 7$.The number of ways to draw $2$ balls out of $7$ is given by ${}^{7}C_{2} = \frac{7 	imes 6}{1 	imes 2} = 21$.So,$n(S) = 21$.Let $E$ be the event that none of the balls drawn is blue. This means the $2$ balls must be chosen from the $2$ red and $3$ green balls,which makes a total of $5$ non-blue balls.The number of ways to draw $2$ balls from these $5$ balls is ${}^{5}C_{2} = \frac{5 	imes 4}{1 	imes 2} = 10$.So,$n(E) = 10$.The probability $P(E) = \frac{n(E)}{n(S)} = \frac{10}{21}$.

$A$ bag contains $2$ red,$3$ green and $2$ blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

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