In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green ?
A. $$\frac{{2}}{{5}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
Answer: Option D
Solution(By Examveda Team)
Total number of balls = (8 + 7 + 6) = 21Let E = event that the ball drawn is neither blue nor green
= Even that the ball drawn is red
∴ n(E) = 8
P(E) = $$\frac{{8}}{{21}}$$
This Question Belongs to Arithmetic Ability >> Probability
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