Question

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 red nor green?

• A. True
• B. False
• C. $\frac{5}{21}$
• D. $\frac{7}{21}$

Answer 1

Answer: (A), (D)

The correct options are
A

True

D $\frac{7}{21}$

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
$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{7}{21}=\frac{1}{3}$