### Related Formula Bayes' Theorem formula: P(E_2|A) = fracP(E_2) cdot P(A|E_2)P(E_1) cdot P(A|E_1) + P(E_2) cdot P(A|E_2) + P(E_3) cdot P(A|E_3)$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2) + P(E_3) \cdot P(A|E_3)}$ ### Core Logic Let the events be defined as: E_1$E_1$: Bag B_1$B_1$ is selected E_2$E_2$: Bag B_2$B_2$ is selected E_3$E_3$: Bag B_3$B_3$ is selected A: The drawn ball is white Since a bag is selected at random: P(E_1) = P(E_2) = P(E_3) = frac13$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$ Conditional probabilities of drawing a white ball from each bag: P(A|E_1) = frac610, quad P(A|E_2) = frac410, quad P(A|E_3) = frac510$P(A|E_1) = \frac{6}{10}, \quad P(A|E_2) = \frac{4}{10}, \quad P(A|E_3) = \frac{5}{10}$ ### Step 1: Substitute and Calculate Substituting the values into Bayes' theorem: P(E_2|A) = fracfrac13 times frac410frac13 times frac610 + frac13 times frac410 + frac13 times frac510$P(E_2|A) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$ P(E_2|A) = frac46 + 4 + 5 = frac415$P(E_2|A) = \frac{4}{6 + 4 + 5} = \frac{4}{15}$ ### Pattern Recognition When bags have equal selection probability, the required conditional probability is simply the number of favorable white balls divided by the total number of white balls across all bags: 4 / (6 + 4 + 5) = 4/15$4 / (6 + 4 + 5) = 4/15$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

### Related Formula Probability of an event P(E') = 1 - P(E)$P(E') = 1 - P(E)$, where E$E$ is the complementary event. ### Core Logic The word GARDEN contains 6 distinct letters: G, A, R, D, E, N. Total number of permutations (words in set S) = 6! = 720$6! = 720$. The vowels present are A and E. In any random arrangement of these letters, there are only 2 possible mutual relative arrangements for the vowels: 1) A appears before E (alphabetical order) 2) E appears before A By symmetry, both relative arrangements are equally probable. ### Step 1: Calculate Probabilities Probability that vowels are in alphabetical order = frac12$\frac{1}{2}$. Therefore, the probability that the selected word will NOT have vowels in alphabetical order is: 1 - frac12 = frac12$1 - \frac{1}{2} = \frac{1}{2}$ ### Pattern Recognition Symmetry Shortcut: For any k$k$ distinct specific objects inside an arrangement of distinct items, the number of ways they can be sorted in a unique relative order is exactly 1/k!$1/k!$ of the total arrangements. Here k=2$k=2$ vowels, so probability of alphabetical order is 1/2! = 1/2$1/2! = 1/2$. Not alphabetical is 1 - 1/2 = 1/2$1 - 1/2 = 1/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 12 Mathematics: Probability