In a box, there are 9 red, 7 blue and 6 green balls.
In a box, there are 9 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green?
Answer/Solution
9/22
Steps/Work
Explanation:
Total number of balls = (9 + 7 + 6) = 22
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 9/22.
Answer: Option D
Total number of balls = (9 + 7 + 6) = 22
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 9/22.
Answer: Option D