$A$ 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
If $A$ and $B$ are two events such that $P(A) = \frac{3}{8}$,$P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$,then $P(A|B) = $
Easy
View Solution$A$ bag contains $3$ red and $5$ black balls and a second bag contains $6$ red and $4$ black balls. $A$ ball is drawn from each bag. The probability that one is red and the other is black,is
Medium
View SolutionThe probability that an ordinary or a non-leap year has $53$ Sundays is
Easy
View SolutionFrom a well-shuffled pack of $52$ cards,one card is drawn at random. The probability that the card drawn is an ace is
Easy
View Solution$A$ bag contains $4$ white balls and $2$ black balls. Another bag contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag,then the probability that both the balls are black,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