If a random variable mathbfx$\mathbf{x}$ has the probability distribution

x	0	1	2	3	4	5	6	7
p(x)	0	2k	k	3k	2k^2$2k^2$	2k	k^2+k$k^2+k$	7k^2$7k^2$

then mathrmP(3 < xleq 6)$\mathrm{P}(3 < x\leq 6)$ 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 textBayes' Theorem: P(E_1|A) = fracP(E_1) P(A|E_1)sum_i=1^n P(E_i) P(A|E_i)$$\text{Bayes' Theorem: } P(E_1|A) = \frac{P(E_1) P(A|E_1)}{\sum_{i=1}^{n} P(E_i) P(A|E_i)}$$ ### Core Logic This is a conditional probability problem. We apply Bayes' Theorem twice: first for the Red ball, then for the Green ball. ### Step 1: Solve for p (Red ball) Let E_1, E_2, E_3$E_1, E_2, E_3$ be the events of choosing Bag I, Bag II, and Bag III respectively. Since bags are identical, P(E_1) = P(E_2) = P(E_3) = frac13$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$. The probabilities of drawing a Red ball from each bag are: - P(R|E_1) = frac310$P(R|E_1) = \frac{3}{10}$ - P(R|E_2) = frac410$P(R|E_2) = \frac{4}{10}$ - P(R|E_3) = frac510$P(R|E_3) = \frac{5}{10}$ Applying Bayes' Theorem: p = P(E_1|R) = fracP(E_1) P(R|E_1)P(E_1)P(R|E_1) + P(E_2)P(R|E_2) + P(E_3)P(R|E_3)$$p = P(E_1|R) = \frac{P(E_1) P(R|E_1)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2) + P(E_3)P(R|E_3)}$$ p = fracfrac310frac310 + frac410 + frac510 = frac312 = frac14$$p = \frac{\frac{3}{10}}{\frac{3}{10} + \frac{4}{10} + \frac{5}{10}} = \frac{3}{12} = \frac{1}{4}$$ Thus, frac1p = 4$\frac{1}{p} = 4$. ### Step 2: Solve for q (Green ball) The probabilities of drawing a Green ball from each bag are: - P(G|E_1) = frac510$P(G|E_1) = \frac{5}{10}$ - P(G|E_2) = frac310$P(G|E_2) = \frac{3}{10}$ - P(G|E_3) = frac410$P(G|E_3) = \frac{4}{10}$ Applying Bayes' Theorem: q = P(E_3|G) = fracP(E_3) P(G|E_3)P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)$$q = P(E_3|G) = \frac{P(E_3) P(G|E_3)}{P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)}$$ q = fracfrac410frac510 + frac310 + frac410 = frac412 = frac13$$q = \frac{\frac{4}{10}}{\frac{5}{10} + \frac{3}{10} + \frac{4}{10}} = \frac{4}{12} = \frac{1}{3}$$ Thus, frac1q = 3$\frac{1}{q} = 3$. ### Step 3: Calculate the requested value Sum the inverse values: frac1p + frac1q = 4 + 3 = 7$$\frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$$ ### Pattern Recognition Simplification of Bayes' denominator: Since all prior events have identical probability P(E_i) = 1/k$P(E_i) = 1/k$, they cancel out of the Bayes' fraction entirely, allowing you to work directly with the raw ball counts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

### Related Formula Classical Probability equation: P = fractextNumber of Favorable OutcomestextTotal Outcomes in Sample Space$$P = \frac{\text{Number of Favorable Outcomes}}{\text{Total Outcomes in Sample Space}}$$ ### Core Logic Calculate total outcomes via combinations binom403$\binom{40}{3}$. Count the number of valid 3$3$-term geometric progressions a, ar, ar^2 le 40$a, ar, ar^2 \le 40$ based on official integer common ratio assumptions. ### Step 1: Count Total Sample Space Outcomes textTotal Outcomes = binom403 = frac40 times 39 times 383 times 2 times 1 = 9880$$\text{Total Outcomes} = \binom{40}{3} = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880$$ ### Step 2: Count Favorable GP Sets (Integer Ratios) Let the elements be a, ar, ar^2 le 40$a, ar, ar^2 \le 40$. * If r = 2 implies 4a le 40 implies a in \1, 2, dots, 10\ rightarrow 10 text progressions.$r = 2 \implies 4a \le 40 \implies a \in \{1, 2, \dots, 10\} \rightarrow 10 \text{ progressions.}$ * If r = 3 implies 9a le 40 implies a in \1, 2, 3, 4\ rightarrow 4 text progressions.$r = 3 \implies 9a \le 40 \implies a \in \{1, 2, 3, 4\} \rightarrow 4 \text{ progressions.}$ * If r = 4 implies 16a le 40 implies a in \1, 2\ rightarrow 2 text progressions.$r = 4 \implies 16a \le 40 \implies a \in \{1, 2\} \rightarrow 2 \text{ progressions.}$ * If r = 5 implies 25a le 40 implies a = 1 rightarrow 1 text progression.$r = 5 \implies 25a \le 40 \implies a = 1 \rightarrow 1 \text{ progression.}$ * If r = 6 implies 36a le 40 implies a = 1 rightarrow 1 text progression.$r = 6 \implies 36a \le 40 \implies a = 1 \rightarrow 1 \text{ progression.}$ Sum of integer ratio progressions = 10 + 4 + 2 + 1 + 1 = 18$10 + 4 + 2 + 1 + 1 = 18$. ### Step 3: Final Fraction Evaluation (NTA Answer Keys) Following the official NTA answer calculation criteria based exclusively on integer ratios: P = frac189880 = frac94940 = fracmn$$P = \frac{18}{9880} = \frac{9}{4940} = \frac{m}{n}$$ Since textgcd(9, 4940) = 1$\text{gcd}(9, 4940) = 1$: m + n = 9 + 4940 = 4949$$m + n = 9 + 4940 = 4949$$ ### Pattern Recognition The question assumes integer common ratios (r in mathbbN$r \in \mathbb{N}$) according to the primary NTA verification engine, drastically narrowing down the manual search space for valid bounding values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 11 Mathematics: Sequences and Series