Question 22 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 14, 2018 by Teachoo

Bag I contains 1 white, 2 black and 3 red balls; Bag II contains 2 white, 1 black and 1 red balls; Bag III contains 4 white, 3 black and 2 red balls. A bag is chosen at random and two balls are drawn from it with replacement. They happen to be one white and one red. What is the probability that they came from Bag III.

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

Question 22
Bag I contains 1 white, 2 black and 3 red balls; Bag II contains 2 white, 1 black and 1 red balls; Bag III contains 4 white, 3 black and 2 red balls. A bag is chosen at random and two balls are drawn from it with replacement. They happen to be one white and one red. What is the probability that they came from Bag III.
Let
A : Event of drawing white ball and red ball
E1 : Event that Bag I is chosen
E2 : Event that Bag II is chosen
E3 : Event that Bag III is chosen
We need to find out the probability that the red and white ball are from Bag III
i.e. P(E3|A)
P(E3|A) = (๐(๐ธ_3 ).๐(๐ด|๐ธ_3))/(๐(๐ธ_1 ).๐(๐ด|๐ธ_1)+๐(๐ธ_2 ).๐(๐ด|๐ธ_2)+๐(๐ธ_3 ).๐(๐ด|๐ธ_3))
"P(E1)" = Probability of Bag I
= 1/3
P(A|E1) = Probability of drawing red and white ball,
if Bag I is selected
Since red can be selected, then white & white can be selected then red
= P(red) ร P(white) + P(white) ร P(red)
= 3/6 ร 1/6 + 1/6 ร 3/6 = ๐/๐
"P(E2)" = Probability of Bag II
= 1/3
P(A|E2) = Probability of drawing red and white ball,
if Bag II is selected
Since red can be selected, then white & white can be selected then red
= P(red) ร P(white) + P(white) ร P(red)
= 1/4 ร 2/4 + 2/4 ร 1/4 = ๐/๐
"P(E3)" = Probability of Bag III
= 1/3
P(A|E2) = Probability of drawing red and white ball,
if Bag III is selected
Since red can be selected, then white & white can be selected then red
= P(red) ร P(white) + P(white) ร P(red)
= 2/9 ร 4/9 + 4/9 ร 2/9 = ๐๐/๐๐
P(E3|A) = (๐(๐ธ_3 ).๐(๐ด|๐ธ_3))/(๐(๐ธ_1 ).๐(๐ด|๐ธ_1)+๐(๐ธ_2 ).๐(๐ด|๐ธ_2)+๐(๐ธ_3 ).๐(๐ด|๐ธ_3))
= (1/3 ร 16/81 )/(1/3 ร 1/6 + 1/3 ร 1/4 + 1/3 ร 16/81)
= (1/3 ร 16/81 )/(1/3 ( 1/6 + 1/4 + 16/81) )
= (16/81 )/(( 1/6 + 1/4 + 16/81) )
= (16/81 )/(( 1/(3 ร 2) + 1/(2 ร 2) + 16/(3 ร 27)) )
= (16/81 )/(( (27 ร 2 + 27 ร 3 + 16 ร 2 ร 2)/(3 ร 27 ร 2 ร 2) ) )
= 16/81 ร (3 ร 27 ร 2 ร 2)/(27 ร 2 ร 2 + 27 ร 3 + 16 ร 2 ร 2)
= 16/81 ร (81 ร 4)/(54 + 81 + 64)
= 16 ร 4/199
= ๐๐/๐๐๐

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.