In a box,there are 8 red,7 blue and 6 green b | vedclass

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(A) Total number of balls in the box $= 8 + 7 + 6 = 21$.Let $S$ be the sample space,so the total number of possible outcomes is $n(S) = 21$.We need to

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$\frac{1}{3}$

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(A) Total number of balls in the box $= 8 + 7 + 6 = 21$.Let $S$ be the sample space,so the total number of possible outcomes is $n(S) = 21$.We need to find the probability that the ball is neither red nor green.If the ball is neither red nor green,it must be blue.The number of blue balls is $7$.Let $E$ be the event of picking a blue ball,so $n(E) = 7$.The probability $P(E)$ is given by $\frac{n(E)}{n(S)} = \frac{7}{21} = \frac{1}{3}$.

In a box,there are $8$ red,$7$ blue and $6$ green balls. One ball is picked up at random. What is the probability that it is neither red nor green?

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