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 textSum of probabilities: sum P(X=x_i) = 1$$\text{Sum of probabilities: } \sum P(X=x_i) = 1$$ textVariance formula: sigma^2 = E(X^2) - mu^2 implies E(X^2) = sigma^2 + mu^2 = 2$$\text{Variance formula: } \sigma^2 = E(X^2) - \mu^2 \implies E(X^2) = \sigma^2 + \mu^2 = 2$$ E(X^2) = sum x_i^2 P(x_i)$$E(X^2) = \sum x_i^2 P(x_i)$$ ### Core Logic Given Probability Distribution: ### Step 1: Set up variance equation We know sigma^2 + mu^2 = sum x_i^2 P(x_i) = 2$\sigma^2 + \mu^2 = \sum x_i^2 P(x_i) = 2$. 0^2left(frac2a+130right) + 1^2left(frac8a-130right) + 2^2left(frac4a+130right) + 3^2(b) = 2$$0^2\left(\frac{2a+1}{30}\right) + 1^2\left(\frac{8a-1}{30}\right) + 2^2\left(\frac{4a+1}{30}\right) + 3^2(b) = 2$$ frac8a-130 + frac16a+430 + 9b = 2$$\frac{8a-1}{30} + \frac{16a+4}{30} + 9b = 2$$ frac24a+330 + 9b = 2$$\frac{24a+3}{30} + 9b = 2$$ 24a + 270b + 3 = 60 implies 24a + 270b = 57$$24a + 270b + 3 = 60 \implies 24a + 270b = 57$$ Dividing by 3: 8a + 90b = 19 quad dots (1)$$8a + 90b = 19 \quad \dots (1)$$ ### Step 2: Total Probability Equation Sum of all probabilities equals 1: frac2a+130 + frac8a-130 + frac4a+130 + b = 1$$\frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1$$ frac14a+130 + b = 1$$\frac{14a+1}{30} + b = 1$$ 14a + 30b + 1 = 30 implies 14a + 30b = 29 quad dots (2)$$14a + 30b + 1 = 30 \implies 14a + 30b = 29 \quad \dots (2)$$ ### Step 3: Solve the Linear System From (2), multiply by 3: 42a + 90b = 87$42a + 90b = 87$. Subtract (1) from this new equation: (42a + 90b) - (8a + 90b) = 87 - 19$$(42a + 90b) - (8a + 90b) = 87 - 19$$ 34a = 68 implies a = 2$$34a = 68 \implies a = 2$$ Substitute a = 2$a = 2$ back into (1): 8(2) + 90b = 19 implies 16 + 90b = 19 implies 90b = 3 implies b = frac130$$8(2) + 90b = 19 \implies 16 + 90b = 19 \implies 90b = 3 \implies b = \frac{1}{30}$$ We need fracab$\frac{a}{b}$: fracab = frac21/30 = 60$$\frac{a}{b} = \frac{2}{1/30} = 60$$ ### Pattern Recognition Notice that sigma^2 + mu^2$\sigma^2 + \mu^2$ is simply the second moment E(X^2)$E(X^2)$. Avoid calculating mu$\mu$ independently. Create a simultaneous system using E(X^2)=c$E(X^2)=c$ and sum p=1$\sum p=1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

### Related Formula P(textEvent) = 1 - P(textComplement of Event)$$P(\text{Event}) = 1 - P(\text{Complement of Event})$$ ### Core Logic For the product ab$ab$ to be divisible by 3, at least one of the numbers a$a$ or b$b$ must be a multiple of 3. It is easier to find the probability of the complement event: neither a$a$ nor b$b$ is a multiple of 3. ### Step 1: Counting Favorable vs Total Outcomes Total numbers = 50. Number of multiples of 3 in the set \1, 2, ldots, 50\$\{1, 2, \ldots, 50\}$ is lfloor frac503 rfloor = 16$\lfloor \frac{50}{3} \rfloor = 16$. Numbers that are NOT multiples of 3 = 50 - 16 = 34$50 - 16 = 34$. The probability that both chosen numbers are NOT multiples of 3 is the number of ways to choose 2 numbers from the 34, divided by the total ways to choose 2 numbers from 50: P(textNot divisible by 3) = frac^34C_2^50C_2$$P(\text{Not divisible by 3}) = \frac{^{34}C_{2}}{^{50}C_{2}}$$ ^34C_2 = frac34 times 332 = 561$$^{34}C_{2} = \frac{34 \times 33}{2} = 561$$ ^50C_2 = frac50 times 492 = 1225$$^{50}C_{2} = \frac{50 \times 49}{2} = 1225$$ ### Step 2: Final Probability Required probability P(ab text is divisible by 3) = 1 - P(textNot divisible by 3)$P(ab \text{ is divisible by 3}) = 1 - P(\text{Not divisible by 3})$ P = 1 - frac5611225 = frac1225 - 5611225 = frac6641225$$P = 1 - \frac{561}{1225} = \frac{1225 - 561}{1225} = \frac{664}{1225}$$ ### Pattern Recognition Whenever a probability question asks for 'at least one' condition (like a product being divisible by a prime), always use the complement rule: 1 - P(textnone)$1 - P(\text{none})$. It transforms a complex multi-case problem into a single combination calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability Class 11 Maths: Permutations and Combinations

### Related Formula sum P(x_i) = 1$$\sum P(x_i) = 1$$ ### Core Logic For a valid probability distribution, the sum of all probabilities must equal 1. 0 + 2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1$$0 + 2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1$$ Combine like terms: 10k^2 + 9k - 1 = 0$$10k^2 + 9k - 1 = 0$$ (10k - 1)(k + 1) = 0$$(10k - 1)(k + 1) = 0$$ Since k$k$ cannot be negative (probabilities must be non-negative), we get k = frac110$k = \frac{1}{10}$. ### Step 1: Calculate the Desired Probability We need to find mathrmP(3 < x leq 6)$\mathrm{P}(3 < x \leq 6)$. This includes the probabilities for x = 4, 5, 6$x = 4, 5, 6$. mathrmP(3 < x leq 6) = P(x=4) + P(x=5) + P(x=6)$$\mathrm{P}(3 < x \leq 6) = P(x=4) + P(x=5) + P(x=6)$$ mathrmP(3 < x leq 6) = 2k^2 + 2k + (k^2 + k) = 3k^2 + 3k$$\mathrm{P}(3 < x \leq 6) = 2k^2 + 2k + (k^2 + k) = 3k^2 + 3k$$ Substitute k = frac110$k = \frac{1}{10}$: = 3left(frac1100right) + 3left(frac110right) = 0.03 + 0.3 = 0.33$$= 3\left(\frac{1}{100}\right) + 3\left(\frac{1}{10}\right) = 0.03 + 0.3 = 0.33$$ ### Pattern Recognition Sum of probabilities always equals 1. This generates a standard quadratic in k$k$. Always discard the negative root since P(x_i) geq 0$P(x_i) \geq 0$ for all i$i$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

### Core Logic Total ways to select a$a$ and b$b$ from 100$100$ natural numbers (ordered and without replacement) is 100 times 99$100 \times 99$. We require a - b geq 10$a - b \geq 10$, which implies a geq b + 10$a \geq b + 10$. ### Step 1: Count Favorable Outcomes Iterate over possible values of a$a$: If a = 100$a = 100$, b$b$ can be anything from 1$1$ to 90$90$ (90$90$ cases). If a = 99$a = 99$, b$b$ can be anything from 1$1$ to 89$89$ (89$89$ cases). ... If a = 11$a = 11$, b$b$ can only be 1$1$ (1$1$ case). Total favorable cases = 1 + 2 + 3 + dots + 90$= 1 + 2 + 3 + \dots + 90$. Using sum of first n$n$ natural numbers: N_textfav = frac90 times 912$$N_{\text{fav}} = \frac{90 \times 91}{2}$$ ### Step 2: Calculate Probability Probability P = fracN_textfavN_texttotal = fracfrac90 times 912100 times 99$P = \frac{N_{\text{fav}}}{N_{\text{total}}} = \frac{\frac{90 \times 91}{2}}{100 \times 99}$ P = frac90 times 912 times 100 times 99 = frac912 times 10 times 11 = frac91220$$P = \frac{90 \times 91}{2 \times 100 \times 99} = \frac{91}{2 \times 10 \times 11} = \frac{91}{220}$$ ### Step 3: Final Answer Here m = 91$m = 91$ and n = 220$n = 220$. They are coprime (gcd is 1). m + n = 91 + 220 = 311$$m + n = 91 + 220 = 311$$ ### Pattern Recognition Ordering constraints like a-bgeq k$a-b\geq k$ on a discrete uniform sample space always collapse into a simple arithmetic progression sum. Start explicitly from the highest/lowest valid bounds to quickly identify the series length. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Probability Class 11 Maths: Sequences and Series

### Related Formula Total Probability Theorem: P(E) = P(E|A_1)P(A_1) + P(E|A_2)P(A_2)$$P(E) = P(E|A_1)P(A_1) + P(E|A_2)P(A_2)$$ ### Core Logic

Let Event W_B$W_B$ be drawing a white ball from Bag B and mixing it in Bag A. Let Event B_B$B_B$ be drawing a black ball from Bag B and mixing it in Bag A. Probability of drawing White from Bag B: P(W_B) = frac610 = frac35$$P(W_B) = \frac{6}{10} = \frac{3}{5}$$ Probability of drawing Black from Bag B: P(B_B) = frac410 = frac25$$P(B_B) = \frac{4}{10} = \frac{2}{5}$$ ### Step 1: Calculating Final Probability If a white ball is transferred, Bag A now has 10 White and 8 Black balls (18 total). Probability of drawing White from A given W_B$W_B$: P(W_A | W_B) = frac1018$$P(W_A | W_B) = \frac{10}{18}$$ If a black ball is transferred, Bag A now has 9 White and 9 Black balls (18 total). Probability of drawing White from A given B_B$B_B$: P(W_A | B_B) = frac918$$P(W_A | B_B) = \frac{9}{18}$$ Total probability of drawing a white ball from Bag A: P(W_A) = P(W_A | W_B)P(W_B) + P(W_A | B_B)P(B_B)$$P(W_A) = P(W_A | W_B)P(W_B) + P(W_A | B_B)P(B_B)$$ P(W_A) = left(frac1018right)left(frac35right) + left(frac918right)left(frac25right)$$P(W_A) = \left(\frac{10}{18}\right)\left(\frac{3}{5}\right) + \left(\frac{9}{18}\right)\left(\frac{2}{5}\right)$$ P(W_A) = frac3090 + frac1890 = frac4890 = frac815$$P(W_A) = \frac{30}{90} + \frac{18}{90} = \frac{48}{90} = \frac{8}{15}$$ ### Step 2: Final Calculation We have P = fracpq = frac815$P = \frac{p}{q} = \frac{8}{15}$. Since gcd(8, 15) = 1$\gcd(8, 15) = 1$, p = 8$p = 8$ and q = 15$q = 15$. p + q = 8 + 15 = 23$$p + q = 8 + 15 = 23$$ ### Pattern Recognition Standard transfer problem. Always set up the exhaustive branches of the initial transfer event and map the new probability state of the target bag. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability