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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.4 Red (R) 4 Black (B) 2 Red (R) 6 Black (B) 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 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)) "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(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 Putting values in formula, P(B1 |R) = (𝟏/𝟐 × 𝟏/𝟐)/(𝟏/𝟐 × 𝟏/𝟒 + 𝟏/𝟐 × 𝟏/𝟐) = (1/4 )/(1/8 + 1/4 ) = ( 2/8 )/( 3/8 ) = 2/3 Therefore, required probability is 𝟐/𝟑

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo