$A$ bag contains $5$ black balls, $4$ white balls, and $3$ red balls. If a ball is selected at random, the probability that it is a black or red ball is
- A$\frac{1}{3}$
- B$\frac{1}{4}$
- C$\frac{5}{12}$
- D$\frac{2}{3}$
Explore More
Similar Questions
Twenty tickets are marked with the numbers $1, 2, \dots, 20$. If three tickets are drawn at random,what is the probability that those marked $7$ and $11$ are among them?
Easy
View SolutionIn a box,there are $8$ red,$7$ blue,and $6$ green balls. One ball is picked up at random. What is the probability that it is neither red nor green?
Easy
View SolutionIn a single throw of $2$ dice,find the probability of getting a total of $12$.
Easy
View SolutionIf $A$ and $B$ are two events such that $A \subseteq B,$ then $P\left( \frac{B}{A} \right) = $
Easy
View SolutionThe corners of regular tetrahedrons are numbered $1, 2, 3, 4$. Three tetrahedrons are tossed. The probability that the sum of the upward corners will be $5$ is
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo