Balls

The total number of balls is $8+7+6=21$

The probability that it is either red or green is $P=\dfrac{8}{21}+\dfrac{6}{21}=\dfrac{2}{3}$

Therefore, the probability that it is neither red nor green is $1-\dfrac{2}{3}=$ $\color{#D61F06}\large \boxed{\dfrac{1}{3}}$

If the ball is neither red nor green it must be blue. There are 21 balls in total of which 7 are blue. Therefore the probability of picking a blue ball is 7/21 which simplifies to 1/3.

Total number of balls = (8 + 7 + 6) = 21.

total green and red balls=8+6=14

so, the probability is (21-14)/21=3

Probability that it is neither red nor green is 1-(8/21 +6/21) = 1/3

The ball must be blue.so,7/total number of ball=7/21=1/3

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue. n(E) = 7.

P(E) = n(E) = 7 = 1 . n(S) 21 3

if the ball is neither red or green, it must be blue and the number of blue balls is 7 so $\frac{7}{21}$ which equals to one third

$P_{red} =$ $\frac{8}{21}$

$P_{green} =$ $\frac{6}{21}$

$P_{red or green} =$ $\frac{8}{21}+\frac{6}{21}=$ $\frac{14}{21}$

$P_{not red or not green} =1-$ $\frac{14}{21}=$ $\frac{1}{3}$

Since total no of balls are 21mand 7 balls are blue therefore probability while picking one ball which is neigher red or green is 7/21 =1/3 Ans K.K.GARg,India