A box contains 4 red balls,5 green balls and | vedclass

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(B) The total number of balls in the box $= 4 + 5 + 6 = 15$.The total number of possible outcomes $n(S) = 15$.We want to find the probability of drawi

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$\frac{3}{5}$

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(B) The total number of balls in the box $= 4 + 5 + 6 = 15$.The total number of possible outcomes $n(S) = 15$.We want to find the probability of drawing either a red or a green ball.The number of red balls $= 4$.The number of green balls $= 5$.The number of favorable outcomes $n(E) = 4 + 5 = 9$.The probability $P(E)$ is given by the ratio of favorable outcomes to total outcomes:$P(E) = \frac{n(E)}{n(S)} = \frac{9}{15}$.Simplifying the fraction by dividing both numerator and denominator by $3$,we get:$P(E) = \frac{3}{5}$.

$A$ box contains $4$ red balls,$5$ green balls and $6$ white balls. $A$ ball is drawn at random from the box. What is the probability that the ball drawn is either red or green?

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