Bag $A$ contains $2$ white and $3$ red balls. Bag $B$ contains $4$ white and $5$ red balls. One bag is chosen at random and a ball is drawn from it,which is found to be red. Find the probability that it was drawn from bag $B$.

  • A
    $5/14$
  • B
    $5/16$
  • C
    $5/18$
  • D
    $25/52$

Explore More

Similar Questions

Let $A, B,$ and $C$ be three mutually independent events. Consider the two statements $S_1$ and $S_2$:
$S_1 : A$ and $B \cup C$ are independent.
$S_2 : A$ and $B \cap C$ are independent.
Then:

There are two dice $A$ and $B$. Die $A$ has $4$ red and $2$ white faces,and die $B$ has $2$ red and $4$ white faces. $A$ coin is tossed once. If it shows head,die $A$ is rolled; if it shows tail,die $B$ is rolled. If the probability that die $A$ is used is $\left(\frac{32}{33}\right)$ given that red turns up every time in the first $n$ throws,then $n=$

$A$ survey of people in a given region showed that $20 \%$ were smokers. The probability of death due to lung cancer given that a person smoked was $10$ times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is $0.006$, what is the probability of death due to lung cancer given that a person is a smoker?

$A$ bag contains $19$ unbiased coins and one coin with head on both sides. One coin is drawn at random and tossed,and a head turns up. If the probability that the drawn coin was unbiased is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $n^2 - m^2$ is equal to:

Recent studies suggest that $12\%$ of the world population is left-handed. Depending on parents' hand usage, the chances of having left-handed children are as follows: $A$. Both parents are left-handed, chances of having left-handed children = $24\%$. $B$. Both parents are right-handed, chance of having left-handed children = $9\%$. $C$. Father left-handed and mother right-handed, chances of having left-handed children = $17\%$. $D$. Father right-handed and mother left-handed, chances of having left-handed children = $22\%$. Given $P(A) = P(B) = P(C) = P(D) = 1/4$ and $L$ denotes the event that the child is left-handed. What is the probability $P(A|L)$?

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