Three distinct numbers are selected randomly from the set \1, 2, 3, dots, 40\$\{1, 2, 3, \dots, 40\}$. If the probability that the selected numbers are in an increasing geometric progression is fracmn$\frac{m}{n}$ where textgcd(m, n) = 1$\text{gcd}(m, n) = 1$, then m + n$m + n$ is equal to ________.

### 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

### Related Formula According to Bayes' Theorem, the conditional probability of an event E_k$E_k$ given that event A$A$ has occurred is: P(E_k|A) = fracP(E_k) cdot P(A|E_k)sum_i=1^n P(E_i) cdot P(A|E_i)$$P(E_k|A) = \frac{P(E_k) \cdot P(A|E_k)}{\sum_{i=1}^{n} P(E_i) \cdot P(A|E_i)}$$ ### Core Logic Let A$A$ be the event that 2$2$ white and 2$2$ black balls are drawn from the bag containing 8$8$ balls. Since 2$2$ white and 2$2$ black balls have already been drawn, the initial composition of the bag could only be one of the following configurations: - E_1$E_1$: 2$2$ White, 6$6$ Black - E_2$E_2$: 3$3$ White, 5$5$ Black - E_3$E_3$: 4$4$ White, 4$4$ Black (Equal composition) - E_4$E_4$: 5$5$ White, 3$3$ Black - E_5$E_5$: 6$6$ White, 2$2$ Black Assuming these 5$5$ configurations are equally likely initially: P(E_1) = P(E_2) = P(E_3) = P(E_4) = P(E_5) = frac15$$P(E_1) = P(E_2) = P(E_3) = P(E_4) = P(E_5) = \frac{1}{5}$$ ### Step 1: Compute Conditional Probabilities We calculate the probability of drawing 2$2$ white and 2$2$ black balls under each hypothesis using combinations: - For E_1$E_1$ (2textW, 6textB$2\text{W}, 6\text{B}$): P(A|E_1) = frac^2C_2 cdot ^6C_2^8C_4 = frac1 cdot 1570 = frac1570$P(A|E_1) = \frac{^{2}C_{2} \cdot ^{6}C_{2}}{^{8}C_{4}} = \frac{1 \cdot 15}{70} = \frac{15}{70}$ - For E_2$E_2$ (3textW, 5textB$3\text{W}, 5\text{B}$): P(A|E_2) = frac^3C_2 cdot ^5C_2^8C_4 = frac3 cdot 1070 = frac3070$P(A|E_2) = \frac{^{3}C_{2} \cdot ^{5}C_{2}}{^{8}C_{4}} = \frac{3 \cdot 10}{70} = \frac{30}{70}$ - For E_3$E_3$ (4textW, 4textB$4\text{W}, 4\text{B}$): P(A|E_3) = frac^4C_2 cdot ^4C_2^8C_4 = frac6 cdot 670 = frac3670$P(A|E_3) = \frac{^{4}C_{2} \cdot ^{4}C_{2}}{^{8}C_{4}} = \frac{6 \cdot 6}{70} = \frac{36}{70}$ - For E_4$E_4$ (5textW, 3textB$5\text{W}, 3\text{B}$): P(A|E_4) = frac^5C_2 cdot ^3C_2^8C_4 = frac10 cdot 370 = frac3070$P(A|E_4) = \frac{^{5}C_{2} \cdot ^{3}C_{2}}{^{8}C_{4}} = \frac{10 \cdot 3}{70} = \frac{30}{70}$ - For E_5$E_5$ (6textW, 2textB$6\text{W}, 2\text{B}$): P(A|E_5) = frac^6C_2 cdot ^2C_2^8C_4 = frac15 cdot 170 = frac1570$P(A|E_5) = \frac{^{6}C_{2} \cdot ^{2}C_{2}}{^{8}C_{4}} = \frac{15 \cdot 1}{70} = \frac{15}{70}$ ### Step 2: Apply Bayes' Theorem We want to find P(E_3|A)$P(E_3|A)$, the probability that the bag contains equal numbers of white and black balls: P(E_3|A) = fracP(E_3) cdot P(A|E_3)sum_i=1^5 P(E_i) cdot P(A|E_i)$$P(E_3|A) = \frac{P(E_3) \cdot P(A|E_3)}{\sum_{i=1}^{5} P(E_i) \cdot P(A|E_i)}$$ P(E_3|A) = fracfrac15 cdot frac3670frac15 left( frac1570 + frac3070 + frac3670 + frac3070 + frac1570 right)$$P(E_3|A) = \frac{\frac{1}{5} \cdot \frac{36}{70}}{\frac{1}{5} \left( \frac{15}{70} + \frac{30}{70} + \frac{36}{70} + \frac{30}{70} + \frac{15}{70} \right)}$$ P(E_3|A) = frac3615 + 30 + 36 + 30 + 15 = frac36126 = frac27$$P(E_3|A) = \frac{36}{15 + 30 + 36 + 30 + 15} = \frac{36}{126} = \frac{2}{7}$$ ### Pattern Recognition Sees: Total number of balls is known, and a sample outcome is given to find the initial state distribution. Shortcut: Notice the symmetry in the configuration possibilities (E_1$E_1$ and E_5$E_5$ have identical probabilities, as do E_2$E_2$ and E_4$E_4$). Sum the numerator terms directly without writing out the full expansion to save crucial seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

### Related Formula n(A cup B cup C) = n(A) + n(B) + n(C) - n(A cap B) - n(B cap C) - n(A cap C) + n(A cap B cap C)$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$$ ### Core Logic Let total sample space S = \1, 2, dots, 50\ implies n(S) = 50$S = \{1, 2, \dots, 50\} \implies n(S) = 50$. * Let A$A$ be multiple of 4: 4, 8, dots, 48 implies n(A) = lfloor frac504 rfloor = 12$4, 8, \dots, 48 \implies n(A) = \lfloor \frac{50}{4} \rfloor = 12$ * Let B$B$ be multiple of 6: 6, 12, dots, 48 implies n(B) = lfloor frac506 rfloor = 8$6, 12, \dots, 48 \implies n(B) = \lfloor \frac{50}{6} \rfloor = 8$ * Let C$C$ be multiple of 7: 7, 14, dots, 49 implies n(C) = lfloor frac507 rfloor = 7$7, 14, \dots, 49 \implies n(C) = \lfloor \frac{50}{7} \rfloor = 7$ ### Step 1: Finding Intersections * A cap B$A \cap B$ (LCM of 4 and 6 = 12): multiples of 12 implies n(A cap B) = lfloor frac5012 rfloor = 4$\implies n(A \cap B) = \lfloor \frac{50}{12} \rfloor = 4$ * B cap C$B \cap C$ (LCM of 6 and 7 = 42): multiples of 42 implies n(B cap C) = lfloor frac5042 rfloor = 1$\implies n(B \cap C) = \lfloor \frac{50}{42} \rfloor = 1$ * A cap C$A \cap C$ (LCM of 4 and 7 = 28): multiples of 28 implies n(A cap C) = lfloor frac5028 rfloor = 1$\implies n(A \cap C) = \lfloor \frac{50}{28} \rfloor = 1$ * A cap B cap C$A \cap B \cap C$ (LCM of 4, 6, 7 = 84): multiples of 84 implies n(A cap B cap C) = 0$\implies n(A \cap B \cap C) = 0$ ### Step 2: Final Enumeration Applying the Principle of Inclusion-Exclusion: n(A cup B cup C) = 12 + 8 + 7 - 4 - 1 - 1 + 0 = 21$$n(A \cup B \cup C) = 12 + 8 + 7 - 4 - 1 - 1 + 0 = 21$$ Therefore, the probability is: P = frac2150$$P = \frac{21}{50}$$ ### Pattern Recognition To quickly count the number of multiples up to N$N$, use the greatest integer function lfloor fracNtextLCM rfloor$\lfloor \frac{N}{\text{LCM}} \rfloor$. This systematically prevents manual counting blunders. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Probability

### Related Formula P(X=k) = q^k-1p$$P(X=k) = q^{k-1}p$$ P(X ge k) = q^k-1$$P(X \ge k) = q^{k-1}$$ P(A|B) = fracP(A cap B)P(B)$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ ### Core Logic This is a geometric probability distribution. Probability of success (rolling a 6) is p = frac16$p = \frac{1}{6}$, and failure is q = frac56$q = \frac{5}{6}$. Calculate a = P(X=3)$a = P(X=3)$: This means the first two tosses are failures, and the third is a success. a = left(frac56right)^2 left(frac16right) = frac25216$$a = \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right) = \frac{25}{216}$$ ### Step 1: Evaluating Cumulative Probability b Calculate b = P(X ge 3)$b = P(X \ge 3)$: This means the first two tosses must be failures (what happens after doesn't matter). b = left(frac56right)^2 = frac2536$$b = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$$ ### Step 2: Evaluating Conditional Probability c Calculate c = P(X ge 6 | X > 3)$c = P(X \ge 6 | X > 3)$: By conditional probability formula: c = fracP(X ge 6 cap X ge 4)P(X ge 4) = fracP(X ge 6)P(X ge 4)$$c = \frac{P(X \ge 6 \cap X \ge 4)}{P(X \ge 4)} = \frac{P(X \ge 6)}{P(X \ge 4)}$$ Using the logic from b$b$: P(X ge 6) = left(frac56right)^5$$P(X \ge 6) = \left(\frac{5}{6}\right)^5$$ P(X ge 4) = left(frac56right)^3$$P(X \ge 4) = \left(\frac{5}{6}\right)^3$$ c = frac(5/6)^5(5/6)^3 = left(frac56right)^2 = frac2536$$c = \frac{(5/6)^5}{(5/6)^3} = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$$ ### Step 3: Final Computation Now compute the requested value fracb+ca$\frac{b+c}{a}$: fracb+ca = fracfrac2536 + frac2536frac25216$$\frac{b+c}{a} = \frac{\frac{25}{36} + \frac{25}{36}}{\frac{25}{216}}$$ fracb+ca = fracfrac5036frac25216$$\frac{b+c}{a} = \frac{\frac{50}{36}}{\frac{25}{216}}$$ fracb+ca = frac5036 times frac21625 = 2 times 6 = 12$$\frac{b+c}{a} = \frac{50}{36} \times \frac{216}{25} = 2 \times 6 = 12$$ ### Pattern Recognition The geometric distribution is "memoryless." Hence, the conditional probability P(X ge k+m | X > m)$P(X \ge k+m | X > m)$ is identical to the unconditional probability P(X ge k)$P(X \ge k)$. Here, P(X ge 6 | X ge 4)$P(X \ge 6 | X \ge 4)$ directly equals P(X ge 3) = b$P(X \ge 3) = b$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability