Three boxes B1,B2,B3 contains balls with different colour as follows. WhiteBlackRedB1212B2324B3432A die is thrown box B1 is choosen if either 1 (or) 2 turns up box B2 is choosen if 3 (or) 4 turns up and box B3 is choosen if 5 (or) 6 turns up. In this way a box is choosen and a ball is drawn from it. If the ball drawn is found to be red then the probability that it is drawn from box B2

PB1=PB2=PB3=13PEB1=25PEB2=49PEB3=29PB2E=P(B2).PEB2P(B1)PEB1+P(B2)PEB2+P(B3)PEB3=512

Three boxes B1,B2,B3 contains balls with different colour as follows. WhiteBlackRedB1212B2324B3432A die is thrown box B1 is choosen if either 1 (or) 2 turns up box B2 is choosen if 3 (or) 4 turns up and box B3 is choosen if 5 (or) 6 turns up. In this way a box is choosen and a ball is drawn from it. If the ball drawn is found to be red then the probability that it is drawn from box B2

PB1=PB2=PB3=13PEB1=25PEB2=49PEB3=29PB2E=P(B2).PEB2P(B1)PEB1+P(B2)PEB2+P(B3)PEB3=512

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Three boxes $B_{1}, B_{2}, B_{3}$ contains balls with different colour as follows.
White Black Red
B₁ 2 1 2
B₂ 3 2 4
B₃ 4 3 2
A die is thrown box B₁ is choosen if either 1 (or) 2 turns up box B₂ is choosen if 3 (or) 4 turns up and box B₃ is choosen if 5 (or) 6 turns up. In this way a box is choosen and a ball is drawn from it. If the ball drawn is found to be red then the probability that it is drawn from box B₂

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{3}{26}$

$\frac{7}{12}$

$\frac{5}{12}$

$\frac{1}{12}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} P (B_{1}) = P (B_{2}) = P (B_{3}) = \frac{1}{3} \\ P (\frac{E}{B_{1}}) = \frac{2}{5} P (\frac{E}{B_{2}}) = \frac{4}{9} P (\frac{E}{B_{3}}) = \frac{2}{9} \end{array}$
$P (\frac{B_{2}}{E}) = \frac{P (B_{2}) . P (\frac{E}{B_{2}})}{P (B_{1}) P (\frac{E}{B_{1}}) + P (B_{2}) P (\frac{E}{B_{2}}) + P (B_{3}) P (\frac{E}{B_{3}})} = \frac{5}{12}$