### Related Formula textBayes' Theorem: P(E_1 | E) = fracP(E_1)P(E | E_1)P(E_1)P(E | E_1) + P(E_2)P(E | E_2)$\text{Bayes' Theorem: } P(E_1 | E) = \frac{P(E_1)P(E | E_1)}{P(E_1)P(E | E_1) + P(E_2)P(E | E_2)}$ ### Core Logic Let E_1$E_1$ be the event that Bag A is selected, and E_2$E_2$ be the event that Bag B is selected. P(E_1) = P(E_2) = frac12$P(E_1) = P(E_2) = \frac{1}{2}$ Let E$E$ be the event that a white ball is drawn. From Bag A (3 white, 7 red, total 10): P(E | E_1) = frac310$P(E | E_1) = \frac{3}{10}$ From Bag B (3 white, 2 red, total 5): P(E | E_2) = frac35$P(E | E_2) = \frac{3}{5}$ ### Step 1: Calculating the Target Probability We need to find the probability that the ball was drawn from Bag A given it is white, i.e., P(E_1 | E)$P(E_1 | E)$. P(E_1 | E) = fracfrac12 times frac310frac12 times frac310 + frac12 times frac35$P(E_1 | E) = \frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{3}{5}}$ Canceling out frac12$\frac{1}{2}$ from the numerator and the denominator: P(E_1 | E) = fracfrac310frac310 + frac610 = frac33 + 6 = frac39 = frac13$P(E_1 | E) = \frac{\frac{3}{10}}{\frac{3}{10} + \frac{6}{10}} = \frac{3}{3 + 6} = \frac{3}{9} = \frac{1}{3}$ ### Pattern Recognition Reverse probability with disjoint prior states directly signals Bayes' theorem. Canceling prior probability terms (P(E_1)=P(E_2)$P(E_1)=P(E_2)$) speeds up the calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability