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

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Question

(A) Total number of balls $= 2 + 3 + 2 = 7$.Let $S$ be the sample space.Then,$N(S) =$ Number of ways of drawing $2$ balls out of $7 = {}^{7}C_{2} = \f

Vedclass · Accepted Answer

$\frac{10}{21}$

Vedclass · Answer

(A) Total number of balls $= 2 + 3 + 2 = 7$.Let $S$ be the sample space.Then,$N(S) =$ Number of ways of drawing $2$ balls out of $7 = {}^{7}C_{2} = \frac{7 	imes 6}{2 	imes 1} = 21$.Let $E$ be the event of drawing two balls,none of which is blue.This means the $2$ balls must be chosen from the $(2 + 3) = 5$ non-blue balls (red and green).Therefore,$N(E) =$ Number of ways of drawing $2$ balls out of $5 = {}^{5}C_{2} = \frac{5 	imes 4}{2 	imes 1} = 10$.$P(E) = \frac{N(E)}{N(S)} = \frac{10}{21}$.

$A$ 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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