Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white,

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 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, 29/45, then n is equal to : [2025]

1 4

2 6

3 3

4 5

Ans.
(2)

We have, Bag 1 : {4W, 5B} and Bag 2 : {nW, 3B}

Also, P(W/Bag 2) = $\frac{29}{45}$

$\Rightarrow P (\frac{W}{B_{1}}) \times P (\frac{W}{B_{2}}) + P (\frac{B}{B_{1}}) \times P (\frac{W}{B_{2}}) = \frac{29}{45}$

$\Rightarrow \frac{4}{9} (\frac{n + 1}{n + 4}) + \frac{5}{9} (\frac{n}{n + 4}) = \frac{29}{45} \Rightarrow \frac{4 n + 4 + 5 n}{9 n + 36} = \frac{29}{45}$

$\Rightarrow \frac{9 n + 4}{9 n + 36} = \frac{29}{45} \Rightarrow 405 n + 180 = 261 n + 1044$

$\Rightarrow 144 n = 864 \Rightarrow n = 6$ .

Want Us to Email you About Special Offers & Updates?