Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
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)$\frac{2}{3}$
(b)$\frac{3}{4}$
(c)$\frac{7}{19}$
(d)$\frac{8}{21}$
(e)$\frac{9}{21}$
Answer
Answer (as printed): D
Explanation
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 red. $\therefore \quad n(E)=8$. $\therefore \quad P(E)=\frac{8}{21}$.
Explanation as extracted from the printed page; notation may be imperfect.