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 S_infty = fraca1 - r$$S_{\infty} = \frac{a}{1 - r}$$ Where S_infty$S_{\infty}$ is the sum of an infinite geometric progression, a$a$ is the first term, and r$r$ is the common ratio. ### Core Logic Let success (S$S$) be rolling a 2, and failure (F$F$) be rolling anything else. P(S) = frac16$P(S) = \frac{1}{6}$ P(F) = frac56$P(F) = \frac{5}{6}$ We want the probability that the first success occurs on an even number of throws (2nd, 4th, 6th, dots$\dots$). The sequence of events for success on even throws is: - Success on 2nd throw: F, S$F, S$ - Success on 4th throw: F, F, F, S$F, F, F, S$ - Success on 6th throw: F, F, F, F, F, S$F, F, F, F, F, S$ Writing this as a sum of probabilities: textRequired Probability = P(F)P(S) + P(F)^3 P(S) + P(F)^5 P(S) + dots$$\text{Required Probability} = P(F)P(S) + P(F)^3 P(S) + P(F)^5 P(S) + \dots$$ = left(frac56right)left(frac16right) + left(frac56right)^3left(frac16right) + left(frac56right)^5left(frac16right) + dots$$= \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^5\left(\frac{1}{6}\right) + \dots$$ ### Step 1: Compute Infinite Series Sum This is an infinite geometric series with: First term a = frac56 times frac16 = frac536$a = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ Common ratio r = left(frac56right)^2 = frac2536$r = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$ Applying the sum formula: S_infty = fracfrac5361 - frac2536$$S_{\infty} = \frac{\frac{5}{36}}{1 - \frac{25}{36}}$$ = fracfrac536frac1136$$= \frac{\frac{5}{36}}{\frac{11}{36}}$$ = frac511$$= \frac{5}{11}$$ ### Pattern Recognition For alternating success/failure probabilities P(textEven) = fracq cdot p1 - q^2$P(\text{Even}) = \frac{q \cdot p}{1 - q^2}$ and P(textOdd) = fracp1 - q^2$P(\text{Odd}) = \frac{p}{1 - q^2}$. Knowing this format immediately turns it into a 5-second mental calculation: frac(5/6)(1/6)1 - 25/36 = 5/11$\frac{(5/6)(1/6)}{1 - 25/36} = 5/11$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 11 Mathematics: Sequences and Series

### 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

### Related Formula P(E) = fractextNumber of favorable outcomestextTotal number of possible outcomes$$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$ ### Core Logic Total possible selections for (x, y)$(x, y)$ with replacement from 0, 1, dots, 10${0, 1, \dots, 10}$ is 11 times 11 = 121$11 \times 11 = 121$. We need pairs (x, y)$(x, y)$ such that |x - y| > 5$|x - y| > 5$, which means x - y > 5$x - y > 5$ or y - x > 5$y - x > 5$. ### Step 1: Counting Favorable Cases Assume x < y$x < y$, so we need y - x geq 6$y - x \geq 6$. If x = 0 Rightarrow y in \6, 7, 8, 9, 10\$x = 0 \Rightarrow y \in \{6, 7, 8, 9, 10\}$ (5 ways) If x = 1 Rightarrow y in \7, 8, 9, 10\$x = 1 \Rightarrow y \in \{7, 8, 9, 10\}$ (4 ways) If x = 2 Rightarrow y in \8, 9, 10\$x = 2 \Rightarrow y \in \{8, 9, 10\}$ (3 ways) If x = 3 Rightarrow y in \9, 10\$x = 3 \Rightarrow y \in \{9, 10\}$ (2 ways) If x = 4 Rightarrow y in \10\$x = 4 \Rightarrow y \in \{10\}$ (1 way) If x ge 5$x \ge 5$, there are no possible values for y$y$ strictly greater than x$x$ satisfying the condition. ### Step 2: Total Probability The number of cases for y > x$y > x$ is 5 + 4 + 3 + 2 + 1 = 15$5 + 4 + 3 + 2 + 1 = 15$. By symmetry, the number of cases for x > y$x > y$ is also 15$15$. Total favorable cases = 15 times 2 = 30$= 15 \times 2 = 30$. Required probability = frac30121$= \frac{30}{121}$. ### Pattern Recognition Absolute difference conditions |x - y| > k$|x - y| > k$ on discrete sets cleanly split into symmetric additive series 1+2+...+n$1+2+...+n$. Calculate one half and multiply by 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

### Related Formula P(X=k) = ^nC_k cdot p^k cdot q^n-k$$P(X=k) = ^nC_k \cdot p^k \cdot q^{n-k}$$ ### Core Logic Given P(H) = 2P(T)$P(H) = 2P(T)$. Since P(H) + P(T) = 1$P(H) + P(T) = 1$, we get 2P(T) + P(T) = 1 implies 3P(T) = 1 implies P(T) = frac13$2P(T) + P(T) = 1 \implies 3P(T) = 1 \implies P(T) = \frac{1}{3}$. Then, P(H) = frac23$P(H) = \frac{2}{3}$. The coin is tossed 3 times. We need the probability of getting exactly 2 tails and 1 head. Using binomial probability: P(2T, 1H) = ^3C_2 times (P(T))^2 times (P(H))^1$$P(2T, 1H) = ^3C_2 \times (P(T))^2 \times (P(H))^1$$ = 3 times left(frac13right)^2 times left(frac23right)$$= 3 \times \left(\frac{1}{3}\right)^2 \times \left(\frac{2}{3}\right)$$ = 3 times frac19 times frac23 = frac627 = frac29$$= 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27} = \frac{2}{9}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

### Core Logic Total marbles = 10 + 30 + 20 + 15 = 75$= 10 + 30 + 20 + 15 = 75$. Drawings are made with replacement, so the events are independent. ### Step 1: Probability Calculation Probability of first drawing a red marble: P(R) = frac1075$P(R) = \frac{10}{75}$. Probability of second drawing a white marble: P(W) = frac3075$P(W) = \frac{30}{75}$. Since they are independent: P(R text and W) = frac1075 times frac3075 = frac475$P(R \text{ and } W) = \frac{10}{75} \times \frac{30}{75} = \frac{4}{75}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability