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

Numerical Answer Type:
Enter a numerical value Answer: 4949 to 4949 +4 marks

Solution & Explanation

### Related Formula Classical Probability equation: P = fractextNumber of Favorable OutcomestextTotal Outcomes in Sample Space ### Core Logic Calculate total outcomes via combinations binom403. Count the number of valid 3-term geometric progressions 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 ### Step 2: Count Favorable GP Sets (Integer Ratios) Let the elements be a, ar, ar^2 le 40. * If 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. * If 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. * If r = 6 implies 36a le 40 implies a = 1 rightarrow 1 text progression. Sum of integer ratio progressions = 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 Since textgcd(9, 4940) = 1: m + n = 9 + 4940 = 4949 ### Pattern Recognition The question assumes integer common ratios (r in mathbbN) 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

Reference Study Guides

More Probability Previous-Year Questions — Page 6

Q19 jee_main_2024_31_jan_morning Variance of Random Variable
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable X to be the number of rotten apples in a draw of two apples, the variance of X is
  • A. frac37153
  • B. frac57153
  • C. frac47153
  • D. frac40153

Solution

### Core Logic Total apples = 18 (3 rotten, 15 good). Random variable X = \0, 1, 2\ representing the number of rotten apples. ### Step 1: Probability Distribution P(X = 0) = frac^15C_2^18C_2 = frac105153 P(X = 1) = frac^3C_1 times ^15C_1^18C_2 = frac45153 P(X = 2) = frac^3C_2^18C_2 = frac3153 ### Step 2: Expectation E(X) = 0 times frac105153 + 1 times frac45153 + 2 times frac3153 = frac51153 = frac13 ### Step 3: Variance E(X^2) = 0 times frac105153 + 1 times frac45153 + 4 times frac3153 = frac57153 Var(X) = E(X^2) - (E(X))^2 = frac57153 - left(frac13right)^2 = frac57153 - frac17153 = frac40153 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

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