Bag A contains 3 white, 7 red balls, and bag B contains 3 white, 2 red balls. One bag is selected at random, and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is [2024]

1 $\frac{3}{10}$

2 $\frac{1}{4}$

3 $\frac{1}{9}$

4 $\frac{1}{3}$

Ans.
(4)

Let $E_{1}$ be the event that bag A is selected, $E_{2}$ be the event that bag B is selected and E be the event that white ball is drawn.

$∴$ $P (E_{1}) = P (E_{2}) = \frac{1}{2}, P (\frac{E}{E_{1}}) = \frac{3}{10}, P (\frac{E}{E_{2}}) = \frac{3}{5}$

$∴ Required probability,$

$P (\frac{E_{1}}{E}) = \frac{P (E_{1}) \cdot P (\frac{E}{E_{1}})}{P (E_{1}) \cdot P (\frac{E}{E_{1}}) + P (E_{2}) \cdot P (\frac{E}{E_{2}})}$

$= \frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{3}{5}} = \frac{1}{3}$

Want Us to Email you About Special Offers & Updates?