# Ex 13.3, 2 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Chapter 13 Class 12 Probability (Term 2)

Serial order wise

Last updated at Feb. 15, 2020 by Teachoo

Ex 13.3, 2 A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.Let B1 : ball is drawn from Bag I B2 : ball is drawn from Bag II R : ball is drawn is red We need to find Now, Probability that ball is drawn from Bag I, if ball is red = P(B1 |R) So, P(B1 |R) = (P(𝐵_1 ) . P(𝑅|𝐵_1))/(P(𝐵_1 ) . P(𝑅|𝐵_1)+P(𝐵_2 ) . P(𝑅|𝐵_2)) Putting values in formula, P(B1 |R) = (1/2 × 1/2)/(1/2 × 1/4 + 1/2 × 1/2) "P(B1 )" = Probability that ball is drawn from Bag I = 𝟏/𝟐 "P(R|B1)" = Probability that ball is red, if drawn from Bag I = 4/(4 + 4) = 4/8 = 𝟏/𝟐 "P(B2)" = Probability that ball is drawn from Bag II = 𝟏/𝟐 "P(R|B2)" = Probability that ball is red, if drawn from Bag II = 2/(2 + 6) = 2/8 = 𝟏/𝟒 = (1/4 )/(1/8 + 1/4 ) = ( 2/8 )/( 3/8 ) = 2/3 Therefore, required probability is 𝟐/𝟑