Bayes theorem

Chapter 13 Class 12 Probability (Term 2)
Concept wise

### Transcript

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(π΅_1 ) . P(π|π΅_1))/(P(π΅_1 ) . P(π|π΅_1)+P(π΅_2 ) . P(π|π΅_2)) Putting values in formula, P(B1 |R) = (1/2 Γ 1/2)/(1/2 Γ 1/4 + 1/2 Γ 1/2) "P(B1 )" = Probability that ball is drawn from Bag I = π/π "P(R|B1)" = Probability that ball is red, if drawn from Bag I = 4/(4 + 4) = 4/8 = π/π "P(B2)" = Probability that ball is drawn from Bag II = π/π "P(R|B2)" = Probability that ball is red, if drawn from Bag II = 2/(2 + 6) = 2/8 = π/π = (1/4 )/(1/8 + 1/4 ) = ( 2/8 )/( 3/8 ) = 2/3 Therefore, required probability is π/π