Three urns respectively contain $2$ white and $3$ black,$3$ white and $2$ black,and $1$ white and $4$ black balls. If one ball is drawn from each urn,then the probability that the selection contains $1$ black and $2$ white balls is

  • A
    $\frac{13}{125}$
  • B
    $\frac{37}{125}$
  • C
    $\frac{28}{125}$
  • D
    $\frac{33}{125}$

Explore More

Similar Questions

The probabilities that players $A$ and $B$ of a team are selected for the captaincy for a tournament are $0.6$ and $0.4$,respectively. If $A$ is selected as the captain,the probability that the team wins the tournament is $0.8$ and if $B$ is selected as the captain,the probability that the team wins the tournament is $0.7$. Then the probability,that the team wins the tournament,is:

Consider the system of equations $ax+by=0, cx+dy=0$,where $a, b, c, d \in \{0, 1\}$.
$STATEMENT-1$: The probability that the system of equations has a unique solution is $3/8$.
$STATEMENT-2$: The probability that the system of equations has a solution is $1$.

Three numbers are selected at random from the set ${1, 2, 3, \dots, 8}$ without replacement. Given that the minimum of the selected numbers is $3$ and the maximum is $6$, what is the probability that the third number is $4$ or $5$?

Difficult
View Solution

The probability that a mechanic makes an error while using a machine on the $n$th day is given by $P(E_n) = \frac{1}{2^n}$. If he has operated the machine for $4$ days,the probability that he has not made a mistake on $3$ of the $4$ days is:

If $A$ and $B$ are two events of a random experiment such that $P(\bar{A})=\frac{2}{3}$,$P(B)=\frac{4}{15}$ and $P(A \cap \bar{B})=\frac{1}{5}$,then $\sqrt{195[P(B \mid(A \cup \bar{B}))+P(A \cup B)]} = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo